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NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry – Here are all the NCERT solutions for Class 10 Maths Chapter 8. This solution contains questions, answers, images, explanations of the complete Chapter 8 titled The Fun They Had of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 8 Introduction to Trigonometry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry in one place.

NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

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Class 10
Subject Maths
Book Mathematics
Chapter Number 8
Chapter Name  

Introduction to Trigonometry

NCERT Solutions Class 10 Maths chapter 8 Introduction to Trigonometry

Class 10, Maths chapter 8, Introduction to Trigonometry solutions are given below in PDF format. You can view them online or download PDF file for future use.

Introduction to Trigonometry

Q.1: In \(\triangle\) ABC right angled at B, AB=24 cm, BC=7 m. Determine
(i) \(\sin A, \cos A\)
(ii) \(\sin C, \cos C\)

Ans : \(\begin{array}{l}{\text { Applying Pythagoras theorem for } \Delta A B C, \text { we obtain }} \\ {A C^{2}=A B^{2}+B C^{2}} \\ {=(24 \mathrm{cm})^{2}+(7 \mathrm{cm})^{2}} \\ {=(576+49) \mathrm{cm}^{2}} \\ {=625 \mathrm{cm}^{2}} \\ {\therefore A C=\sqrt{625} \mathrm{cm}=25 \mathrm{cm}}\end{array}\) \(\sin A=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C}\) \(=\frac{7}{25}\) \(\cos A=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{24}{25}\) (ii) \(\begin{array}{l}{ \sin \mathrm{C}=\frac{\text { Side opposite to } \angle \mathrm{C}}{\text { Hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}} \\ {=\frac{24}{25}}\end{array}\) \(\begin{array}{l}{\cos \mathrm{C}=\frac{\text { Side adjacent to } \angle \mathrm{C}}{\mathrm{Hypotenuse}}=\frac{\mathrm{BC}}{\mathrm{AC}}} \\ {=\frac{7}{25}}\end{array}\)

Q.2: In the given figure find tan P - cot R.
 

Ans : \(\begin{array}{l}{\text { Applying Pythagoras theorem for } \Delta P Q R, \text { we obtain }} \\ {P R^{2}=P Q^{2}+Q R^{2}} \\ {(13 \mathrm{cm})^{2}=(12 \mathrm{cm})^{2}+Q R^{2}} \\ {169 \mathrm{cm}^{2}=144 \mathrm{cm}^{2}+Q R^{2}} \\ {25 \mathrm{cm}^{2}=Q R^{2}} \\ {Q R=5 \mathrm{cm}}\end{array}\) \(\begin{aligned} \tan \mathbf{P} &=\frac{\text { Side opposite to } \angle P}{\text { Side adjacent to } \angle P}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \\ \cot R &=\frac{\text { Side adjacent to } \angle R}{\text { Side opposite to } \angle R}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \end{aligned}\) \(\tan \mathrm{P}-\cot \mathrm{R}=\frac{5}{12}-\frac{5}{12}=0\)

Q.3: If \(\sin A=\frac{3}{4}, \text { calculate cos } A \text { and } \tan A.\)

Ans : Let\(\Delta A B C \text { be a right-angled triangle, right-angled at point } B\) Given that, \(\begin{array}{l}{\sin A=\frac{3}{4}} \\ {\frac{B C}{A C}=\frac{3}{4}}\end{array}\) Let BC be 3k . Therefore, AC will be 4k , where k is a positive integer. Applying Pythagoras theorem in \(\triangle ABC \), we obtain \(\begin{array}{l}{A C^{2}=A B^{2}+B C^{2}} \\ {(4 k)^{2}=A B^{2}+(3 k)^{2}} \\ {16 k^{2}-9 k^{2}=A B^{2}} \\ {7 k^{2}=A B^{2}} \\ {A B=\sqrt{7} k}\end{array}\) \(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotent } t 0 \angle A} \\ &=\frac{A B}{A C}=\frac{\sqrt{7 k}}{4 k}=\frac{\sqrt{7}}{4} \\ \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A} \\ &=\frac{B C}{A B}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}} \end{aligned}\)

Q.4: Given \(15 \cot A=8 . \text { Find } \sin A \text { and } \sec A.\)

Ans : Consider a right-angled triangle, right-angled at B. Given that, \(\begin{array}{l}{\sin A=\frac{3}{4}} \\ {\frac{B C}{A C}=\frac{3}{4}}\end{array}\) Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer. Applying Pythagoras theorem in \(\triangle \mathrm{ABC}\) , we obtain \(\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}} \\ {(4 k)^{2}=\mathrm{AB}^{2}+(3 \mathrm{k})^{2}} \\ {16 \mathrm{k}^{2}-\mathrm{gk}^{2}=\mathrm{AB}^{2}} \\ {7 k^{2}=\mathrm{AB}^{2}} \\ {\mathrm{AB}=\sqrt{7} k}\end{array}\) \(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }} \\ &=\frac{A B}{A C}=\frac{\sqrt{7 k}}{4 k}=\frac{\sqrt{7}}{4} \end{aligned}\) \(\begin{aligned} \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A} \\ &=\frac{B C}{A B}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}} \end{aligned}\) \(\begin{aligned} \cot A &=\frac{\text { Side adjacent to } \angle A}{\text { Side opposite to } \angle A} \\ &=\frac{A B}{B C} \end{aligned}\) \(\begin{array}{c}{\text { It is given that, }} \\ {\text { cot } A=\frac{8}{15}} \\ {\frac{A B}{B C}=\frac{8}{15}}\end{array}\) Let AB be 8k.Therefore, 3C will be 15k, where k is a positive integer. Applying Pythagoras theorem in \(\triangle\)ABC, we obtain \(\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}} \\ {=(8 k)^{2}+(15 k)^{2}} \\ {=64 k^{2}+225 \mathrm{k}^{2}} \\ {=289 \mathrm{k}^{2}} \\ {\mathrm{AC}=17 k}\end{array}\) \(\begin{aligned} \sin A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C} \\ &=\frac{15 k}{17 k}=\frac{15}{17} \\ \sec A &=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A} \\ &=\frac{A C}{A B}=\frac{17}{8} \end{aligned}\)

Q.5: Given \(\sec \theta=\frac{13}{12},\) calculate all other trigonometric ratios.

Ans : Consider a right-angle \(\triangle \mathrm{ABC}, \text { right-angled at point } \mathrm{B}\) \(\begin{array}{l}{\sec \theta=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle \theta}} \\ {\frac{13}{12}=\frac{\mathrm{AC}}{\mathrm{AB}}}\end{array}\) If AC is 13k, AB will be 12k, where k is a positive integer. Applying Pythagoras theorem in AABC, we obtain \(\begin{array}{l}{(A C)^{2}=(A B)^{2}+(B C)^{2}} \\ {(13 k)^{2}=(12 k)^{2}+(B C)^{2}} \\ {169 k^{2}=144 k^{2}+B C^{2}} \\ {25 k^{2}=B C^{2}}\end{array}\) BC = 5k \(\sin \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}\) \(\cos \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}\) \(\tan \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Side adjacent to } \angle \theta}=\frac{B C}{A B}=\frac{5 k}{12 k}=\frac{5}{12}\) \(\cot \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Side opposite to } \angle \theta}=\frac{A B}{B C}=\frac{12 k}{5 k}=\frac{12}{5}\) \(cosec \ \theta=\frac{\text { Hypotenuse }}{\text { Side opposite to } \angle \theta}=\frac{A C}{B C}=\frac{13 k}{5 k}=\frac{13}{5}\)

Q.6: \(\begin{array}{l}{\text { If } \angle A \text { and } \angle B \text { are acute angles such that } \cos A=\cos B \text { , then show that }} \\ {\angle A=\angle B}\end{array}\)

Ans : Let us consider a triangle ABC in which \(\mathrm{CD} \perp \mathrm{AB}\) It is given that, \(\begin{array}{l}{\cos A=\cos B} \\ {\Rightarrow \frac{A D}{A C}=\frac{B D}{B C} \ldots}\end{array}\) We have to prove \(\angle A=\angle B\). To prove this, let us extend AC to P such that BC = CP From equation (1) we obtain, \(\begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}} \\ {\Rightarrow \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{CP}}}\end{array} \quad(\text { By construction, we have } \mathrm{BC}=\mathrm{CP})\) By using the converse of B.P.T, \(\begin{array}{l}{\mathrm{CD} \| \mathrm{BP}} \\ {\Rightarrow \angle \mathrm{ACD}=\angle \mathrm{CPB}(\text { Corresponding angles }) \ldots \text { (3) }} \\ {\text { And, } \angle \mathrm{BCD}=\angle \mathrm{CBP} \text { (Alternate interior angles)... (4) }}\end{array}\) \(\begin{array}{l}{\text { By construction, we have } \mathrm{BC}=\mathrm{CP} \text { . }} \\ {\therefore \angle \mathrm{CBP}=\angle \mathrm{CPB} \text { (Angle opposite to equal sides of a triangle)... (5) }} \\ {\text { From equations }(3),(4), \text { and }(5), \text { we obtain }} \\ {\angle \mathrm{ACD}=\angle \mathrm{BCD} \ldots(6)}\end{array}\) \(\begin{array}{l}{\angle C D A=\angle C D B\left[B o t h 90^{\circ}\right]} \\ {\text { Therefore, the remaining angles should be equal. }} \\ {\therefore \angle C A D=\angle C B D} \\ {\Rightarrow \angle A=\angle B}\end{array}\) Alternatively, Let us consider triangle ABC in which \(C D \perp A B\) It is given that, Cos A = Cos B \(\begin{array}{l}{\Rightarrow \frac{A D}{A C}=\frac{B D}{B C}} \\ {\Rightarrow \frac{A D}{B D}=\frac{A C}{B C}} \\ {\text { Let } \frac{A D}{B D}=\frac{A C}{B C}=k} \\ {\Rightarrow A D=k B D \ldots(1)} \\ {\text { And, } A C=k B C \ldots(2)}\end{array}\) using Pythagoras theorem for triangles CAD and CBD, we obtain \(\begin{array}{l}{\mathrm{CD}^{2}=\mathrm{AC}^{2}-\mathrm{AD}^{2} \ldots(3)} \\ {\text { And, } \mathrm{CD}^{2}=\mathrm{BC}^{2}-\mathrm{BD}^{2} \ldots \text { (4) }} \\ {\text { From equations }(3) \text { and }(4), \text { we obtain }} \\ {\mathrm{AC}^{2}-\mathrm{AD}^{2}=\mathrm{BC}^{2}-\mathrm{BD}^{2}}\end{array}\) \(\begin{array}{l}{\Rightarrow(k B C)^{2}-(k B D)^{2}=B C^{2}-B D^{2}} \\ {\Rightarrow k^{2}\left(B C^{2}-B D^{2}\right)=B C^{2}-B D^{2}} \\ {\Rightarrow k^{2}=1} \\ {\Rightarrow k=1}\end{array}\) Putting this value in equation (2), we obtain AC = BC \(\Rightarrow \angle A=\angle B\)(Angles opposite to equal sides of a angle)

Q.7: \(\begin{array}{l}{\text { If } \cot \theta=\frac{7}{8}, \text { evaluate }} \\ { ( i )}{\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\left(\text { (ii) } \cot ^{2} \theta\right.}\end{array}\)

Ans : Let us consider a right triangle ABC, right-angled at point B. \(\begin{aligned} \cot \theta &=\frac{\text { Side adjacent to } \angle \theta}{\text { Side opposite to } \angle \theta}=\frac{\mathrm{BC}}{\mathrm{AB}} \\ &=\frac{7}{8} \end{aligned}\) If BC is 7k, then AB will be 8k, where k is a positive integer. Applying Pythagoras theorem in \(\triangle\)ABC, we obtain \(\begin{array}{l}{A C^{2}=A B^{2}+B C^{2}} \\ {=(8 k)^{2}+(7 k)^{2}} \\ {=64 k^{2}+49 k^{2}} \\ {=113 k^{2}} \\ {A C=\sqrt{113} k}\end{array}\) \(\begin{aligned} \sin \theta &=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}} \\ &=\frac{8 k}{\sqrt{113} k}=\frac{8}{\sqrt{113}} \\ \cos \theta &=\frac{\text { Side adjacent to } \angle \theta}{\text { Hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}} \\ &=\frac{7 k}{\sqrt{113} k}=\frac{7}{\sqrt{113}} \end{aligned}\) (i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{\left(1-\sin ^{2} \theta\right)}{\left(1-\cos ^{2} \theta\right)}\) \(=\frac{1-\left(\frac{8}{\sqrt{113}}\right)^{2}}{1-\left(\frac{7}{\sqrt{113}}\right)^{2}}=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}\) \(=\frac{\frac{49}{113}}{\frac{64}{113}}=\frac{49}{64}\) (ii) \(\cot ^{2} \theta=(\cot \theta)^{2}=\left(\frac{7}{8}\right)^{2}=\frac{49}{64}\)

Q.8: If \(3 \cot A=4, \text { Check whether } \frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A\) or not.

Ans : It is given that 3 Cot A = 4 Or, Cot A = \(\frac{4}{3}\) Consider a right triangle ABC, right-angled at point B. \(\cot A=\frac{\text { Side adjacent to } \angle A}{\text { Side opposite to } \angle A}\) \(\frac{A B}{B C}=\frac{4}{3}\) If AB is 4k , then BC will be 3k, where k is a positive integer. In \(\triangle \mathrm{ABC}\), \(\begin{array}{l}{(A C)^{2}=(A B)^{2}+(B C)^{2}} \\ {=(4 k)^{2}+(3 k)^{2}} \\ {=16 k^{2}+9 k^{2}} \\ {=25 k^{2}} \\ {A C=5 k}\end{array}\) \(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }}=\frac{A B}{A C} \\ &=\frac{4 k}{5 k}=\frac{4}{5} \end{aligned}\) \(\begin{aligned} \sin A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C} \\ &=\frac{3 k}{5 k}=\frac{3}{5} \end{aligned}\) \(\begin{aligned} \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A B} \\ &=\frac{3 k}{4 k}=\frac{3}{4} \end{aligned}\) \(\begin{array}{c}{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}} \\ {=\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}} \end{array}\) \(\begin{array}{l}{\cos ^{2} A-\sin ^{2} A=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}} \\ {=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}} \\ {\therefore \quad \frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A}\end{array}\)

Q.9: \(\begin{array}{l}{\text { In } \triangle A B C, \text { right angled at } B \text { . If } A=\frac{1}{\sqrt{3}}, \text { find the value of }} \\ {\text { (i) } \sin A \cos C+\cos A \sin C} \\ {\text { (ii) } \cos A \cos C-\sin A \sin C}\end{array}\)

Ans : \(\begin{array}{l}{\tan A=\frac{1}{\sqrt{3}}} \\ {\frac{B C}{A B}=\frac{1}{\sqrt{3}}}\end{array}\) If BC is k , then AB will be , where k is a positive integer. In \(\triangle\)ABC, \(\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}} \\ {=(\sqrt{3} k)^{2}+(k)^{2}} \\ {=3 \mathrm{k}^{2}+\mathrm{k}^{2}=4 \mathrm{k}^{2}} \\ {\therefore \mathrm{AC}=2 \mathrm{k}}\end{array}\) \(\sin A=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}\) \(\cos A=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\) \(\sin C=\frac{\text { Side opposite to } \angle C}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\) \(\cos C=\frac{\text { Side adjacent to } \angle C}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}\) (i) sin A cos C + cos A sin C \(\begin{array}{l}{=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4}+\frac{3}{4}} \\ {=\frac{4}{4}=1}\end{array}\) (ii) cos A cos C - sin A sin C \(=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0\)

Q.10: \(\begin{array}{l}{\text { In } \Delta \mathrm{PQR}, \text { right angled at } \mathrm{Q}, \mathrm{PR}+\mathrm{QR}=25 \mathrm{cm} \text { and } \mathrm{PQ}=5 \mathrm{cm} . \text { Determine the values }} \\ {\text { of } \sin \mathrm{P}, \cos \mathrm{P} \text { and } \tan \mathrm{P}}\end{array}\)

Ans : Given that, PR + QR = 25 PQ = 5 Let PR be x. Therefore, QR = 25 -x Applying Pythagoras theorem in \(\Delta\)PQR, we obtain \(\begin{array}{l}{\mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{QR}^{2}} \\ {x^{2}=(5)^{2}+(25-x)^{2}} \\ {x^{2}=25+625+x^{2}-50 x} \\ {50 x=650} \\ {x=13} \\ {\text { Therefore, } \mathrm{PR}=13 \mathrm{cm}} \\ {\mathrm{QR}=(25-13) \mathrm{cm}=12 \mathrm{cm}}\end{array}\) \(\sin \mathrm{P}=\frac{\text { Side opposite to } \angle \mathrm{P}}{\text { Hypotenuse }}=\frac{\mathrm{QR}}{\mathrm{PR}}=\frac{12}{13}\) \(\cos P=\frac{\text { Side adjacent to } \angle P}{\text { Hypotenuse }}=\frac{P Q}{P R}=\frac{5}{13}\) \(\tan P=\frac{\text { Side opposite to } \angle P}{\text { Side adjacent to } \angle P}=\frac{Q R}{P Q}=\frac{12}{5}\)

Q.11: \(\begin{array}{l}{\text { State whether the following are true or false. Justify your answer. }} \\ {\text { (1) The value of tan } A \text { is always less than } 1 \text { . }} \\ {\text { (ii) } \sec A=\frac{12}{5} \text { for some value of angle A. }} \\ {\text { (iii) } \cos A \text { is the abbreviation used for the cosecant of angle A. }}\end{array}\)
(iv) cot A is the product of cot and  A
\((v) \sin \theta=\frac{4}{3}, \text { for some angle } \theta\)

Ans : (i) Consider a \(\Delta\)ABC, right-angled at B. \(\begin{aligned} \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A} \\ &=\frac{12}{5} \end{aligned}\) \(\begin{array}{l}{\text { But } \frac{12}{5}>1} \\ {\therefore \tan A>1} \\ {\text { So, tan } A<1 \text { is not always true. }} \\ {\text { Hence, the given statement is false. }}\end{array}\) (ii) \(\sec A=\frac{12}{5}\) \(\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A}=\frac{12}{5}\) \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{12}{5}\) Let AC be 12k, AB will be 5k, where k is a positive integer. Applying Pythagoras theorem in \(\Delta\) ABC, we obtain \(\begin{array}{l}{A C^{2}=A B^{2}+B C^{2}} \\ {(12 k)^{2}=(5 k)^{2}+B C^{2}} \\ {144 k^{2}=25 k^{2}+B C^{2}} \\ {B C^{2}=119 k^{2}} \\ {B C=10.9 k}\end{array}\) It can be observed that for given two sides AC = 12k and AB = 5k, BC should be such that, AC - AB < BC < AC + AB 12k - 5k < BC < 12k + 5k 7k < BC < 17k However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible. Hence, the given statement is true. (iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false. (iv) cot A is not the product of cot not A. It is the cotangent of \(\angle \)A. Hence, the given statement is false. (v) \(\sin \theta=\frac{4}{3}\) We know that in a right-angled triangle, \(\sin \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}\) In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin \(\theta\) is not possible. Hence, the given statement is false.

Q.1: Evaluate the following 
(i) \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\)
(ii) \(2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}\)
(iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\csc 30^{\circ}}\)
(iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\csc 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)

Ans : (i) \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\) \(\begin{array}{l}{=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)} \\ {=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1}\end{array}\) (ii) \(2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}\) \(\begin{array}{l}{=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ {=2+\frac{3}{4}-\frac{3}{4}=2}\end{array}\) (iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+cosec 30^{\circ}}\) \(\begin{array}{l}{=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}} \\ {=\frac{\sqrt{3}}{\sqrt{2}(2+2 \sqrt{3})}=\frac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}}}\end{array}\) \(\begin{array}{l}{=\frac{\sqrt{3}(2 \sqrt{6}-2 \sqrt{2})}{(2 \sqrt{6}+2 \sqrt{2})(2 \sqrt{6}-2 \sqrt{2})}} \\ {=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(2 \sqrt{6})^{2}-(2 \sqrt{2})^{2}}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{24-8}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{16}} \\ {=\frac{\sqrt{18}-\sqrt{6}}{8}=\frac{3 \sqrt{2}-\sqrt{6}}{8}}\end{array}\) (iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\) \(\begin{array}{l}{=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}} \\ {=\frac{3 \sqrt{3}-4}{\frac{3 \sqrt{3}-4}{2 \sqrt{3}+4}}=\frac{(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)}}\end{array}\) \(\begin{array}{l}{=\frac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}=\frac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}} \\ {=\frac{27+16-24 \sqrt{3}}{27-16}=\frac{43-24 \sqrt{3}}{11}}\end{array}\) (v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\) \(\begin{aligned} &=\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\=& \frac{5\left(\frac{1}{4}\right)+\left(\frac{16}{3}\right)-1}{\frac{1}{4}+\frac{3}{4}} \\=& \frac{15+64-12}{\frac{4}{4}}=\frac{67}{12} \end{aligned}\)

Q.2: Choose the correct option and justify your choice:
\((i) \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\)
\(\begin{array}{lllll}{\text { (A) }} & {\sin 60^{\circ}} & {\text { (B) } \cos 60^{\circ}} & {\text { (C) tan } 60^{\circ}} & {\text { (D) } \sin 30^{\circ}}\end{array}\)
\((ii) \frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\)
\(\text { (A) }\tan 90^{\circ} \quad \text { (B) } 1 \quad \text { (C) } \sin 45^{\circ} \quad \text { (D) } 0\)
\((iii)\sin 2 A=2 \sin A \text { is true when } A=\)
\(\text { (A) }0^{\circ} \quad \text { (B) } 30^{\circ} \quad \text { (C) }  45^{\circ} \quad \text { (D) } 60^{\circ}\)
\((iv)\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\)
\(\begin{array}{lll}{\text { (A) } \cos 60^{\circ}} & {\text { (B) } \sin 60^{\circ}} & {\text { (C) tan } 60^{\circ}} & {\text { (D) } \sin 30^{\circ}}\end{array}\)

Ans : (i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) \(\begin{array}{l}{=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{\sqrt{3}}{\frac{4}{3}}} \\ {=\frac{6}{4 \sqrt{3}}=\frac{\sqrt{3}}{2}}\end{array}\) Out of the given alternatives, only \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\) Hence, (A) is correct. (ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) \(=\frac{1-(1)^{2}}{1+(1)^{2}}=\frac{1-1}{1+1}=\frac{0}{2}=0\) Hence, (D) is correct. (iii) Out of the given alternatives, only A = 0\(^\circ\) is correct. As sin 2A = sin \(^\circ\) = 0 2 sin A = 2 sin \(^\circ\) = 2(0) = 0 Hence, (A) is correct. (iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) \(\begin{aligned} =\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \\= \sqrt{3} \end{aligned}\) Out of the given alternatives, only tan\(60^{\circ}=\sqrt{3}\) Hence, (C) is correct.

Q.3: \(\begin{array}{l}{\text { If } \tan (A+B)=\sqrt{3} \text { and }(A-B)=\frac{1}{\sqrt{3}}} \\ {0^{\circ} < A+B \leq 90^{\circ}, A > B \text { find } A \text { and } B \text { . }}\end{array}\)

Ans : \(\begin{array}{l}{\tan (\mathrm{A}+\mathrm{B})=\sqrt{3}} \\ {\Rightarrow \tan (\mathrm{A}+\mathrm{B})=\tan 60} \\ {\Rightarrow \mathrm{A}+\mathrm{B}=60 \ldots(1)} \\ {\tan (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{3}}} \\ {\Rightarrow \tan (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{3}}} \\ {\Rightarrow \mathrm{A}-\mathrm{B}=30 \ldots \text { (2) }} \\ {\Rightarrow \mathrm{A}-\mathrm{B}=30 \ldots \text { (2) }} \\ {\text { On adding both equations, we obtain }}\end{array}\) \(\begin{array}{l}{2 A=90} \\ {\Rightarrow A=45} \\ {\text { From equation }(1), \text { we obtain }} \\ {45+B=60} \\ {B=15} \\ {\text { Therefore, } \angle A=45^{\circ} \text { and } \angle B=15^{\circ}}\end{array}\)

Q.4: \(\begin{array}{l}{\text { State whether the following are true or false. Justify your answer. }} \\ {\text { (i) } \sin (A+B)=\sin A+\sin B}\end{array}\)
\(\begin{array}{l}{\text { (ii) The value of sin\(\theta\) increases as } \theta \text { increases }} \\ {\text { (iii) The value of } \cos \theta \text { increases as } \theta \text { increases }} \\ {\text { (iv) } \sin \theta=\cos \theta \text { for all values of } \theta} \\ {\text { (v) } \cot A \text { is not defined for } A=0^{\circ}}\end{array}\)

Ans : \(\begin{array}{l}{\text { (i) } \sin (A+B)=\sin A+\sin B} \\ {\text { Let } A=30^{\circ} \text { and } B=60^{\circ}} \\ {\sin (A+B)=\sin \left(30^{\circ}+60^{\circ}\right)} \\ {=\sin 90^{\circ}} \\ {=1}\end{array}\) \(\begin{array}{l}{\sin A+\sin B=\sin 30^{\circ}+\sin 60^{\circ}} \\ {=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}} \\ {\text { Clearly, } \sin (A+B) \neq \sin A+\sin B} \\ {\text { Hence, the given statement is false. }}\end{array}\) \(\begin{array}{l}{\text { (ii) The value of } \sin \theta \text { increases as } \theta \text { increases in the interval of } 0^{\circ}<\theta<90^{\circ} \text { as }} \\ {\sin 0^{\circ}=0} \\ {\sin 30^{\circ}=\frac{1}{2}=0.5} \\ {\sin 45^{\circ}=\frac{1}{\sqrt{2}}=0.707} \\ {\sin 60^{\circ}=\frac{\sqrt{3}}{2}=0.866} \\ {\sin 90^{\circ}=\frac{\sqrt{3}}{2}=0.866} \\ {\text { Hence, the given statement is true. }}\end{array}\) (iii) \(\cos 0^{\circ}=1\)\(\begin{array}{l}{\cos 30^{\circ}=\frac{\sqrt{3}}{2}=0.866} \\ {\cos 45^{\circ}=\frac{1}{\sqrt{2}}=0.707} \\ {\cos 60^{\circ}=\frac{1}{2}=0.5} \\ {\cos 90^{\circ}=0} \\ {\text { It can be observed that the value of } \cos \theta \text { does not increase in the interval of } 0^{\circ}<\theta} \\ {\text { Hence, the given statement is false. }}\end{array}\) \(\begin{array}{l}{\text { (iv) } \sin \theta=\cos \theta \text { for all values of } \theta} \\ {\text { This is true when } \theta=45^{\circ}} \\ {\sin 45^{\circ}=\frac{1}{\sqrt{2}}} \\ {\cos 45^{\circ}=\frac{1}{\sqrt{2}}} \\ {\text { It is not true for all other values of } \theta \text { . }} \\ {\text { Hence, the given statement is false. }}\end{array}\) \(\begin{array}{l}{\text { (v) cot } A \text { is not defined for } A=0^{\circ}} \\ {\cot A=\frac{\cos A}{\sin A}} \\ {\cot 0^{\circ}=\frac{\cos 0^{\circ}}{\sin 0^{\circ}}=\frac{1}{0}=\text { undefined }} \\ {\text { Hence, the given statement is true. }}\end{array}\)

Q.1: \(\begin{array}{l}{\text { Evaluate: }} \\ {\text { (i) } \frac{\sin 18^{\circ}}{\cos 72^{\circ}} \quad \text { (ii) } \frac{\tan 26^{\circ}}{\cot 64^{\circ}} \quad \text { (iii) } \cos 48^{\circ}-\sin 42^{\circ} \quad \text { (iv) } cosec 31^{\circ}-\sec 59^{\circ}}\end{array}\)

Ans : (i) \(\begin{array}{l}{\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}} \\ {=\frac{\cos 72^{\circ}}{\cos 72^{\circ}}=1}\end{array}\) \((ii){\frac{\tan 26^{\circ}}{\cot 64^{\circ}}=\frac{\tan \left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}} \\ {=\frac{\cot 64^{\circ}}{\cot 64^{\circ}}=1} \\ {(\mathrm{iii}) \cos 48^{\circ}-\sin 42^{\circ}=\cos \left(90^{\circ}-42^{\circ}\right)-\sin 42^{\circ}} \\ {=\sin 42^{\circ}-\sin 42^{\circ}} \\ {=0}\) \(\begin{array}{l}{(\mathrm{iv}) \cos \mathrm{ec} 31^{\circ}-\sec 59^{\circ}=cosec \left(90^{\circ}-59^{\circ}\right)-\sec 59^{\circ}} \\ {=\sec 59^{\circ}-\sec 59^{\circ}} \\ {=0}\end{array}\)

Q.2: Show that:
\({\text { (i) } \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1} \\ {\text { (ii) } \cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0}\)

Ans : \(\begin{array}{l}{\text { (i) tan } 48^{\circ} \text { tan } 23^{\circ} \text { tan } 42^{\circ} \text { tan } 67^{\circ}} \\ {=\tan \left(90^{\circ}-42^{\circ}\right) \tan \left(90^{\circ}-67^{\circ}\right) \tan 42^{\circ} \tan 67^{\circ}} \\ {=\cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}} \\ {=\left(\cot 42^{\circ} \tan 42^{\circ}\right)\left(\cot 67^{\circ} \tan 67^{\circ}\right)} \\ {=(1)(1)} \\ {=1}\end{array}\) \(\begin{array}{l}{\text { (ii) } \cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}} \\ {=\cos \left(90^{\circ}-52^{\circ}\right) \cos \left(90^{\circ}-38^{\circ}\right)-\sin 38^{\circ} \sin 52^{\circ}} \\ {=\sin 52^{\circ} \sin 38^{\circ}-\sin 38^{\circ} \sin 52^{\circ}} \\ {=0}\end{array}\)

Q.3: \({\text{If tan }}2A=\cot \left(A-18^{\circ}\right), \text { where } 2 A \text { is an acute angle, find the value of } A.\)

Ans : \(\begin{array}{l}{\text { Given that, }} \\ {\tan 2 A=\cot \left(A-18^{\circ}\right)} \\ {\cot \left(90^{\circ}-2 A\right)=\cot \left(A-18^{\circ}\right)} \\ {90^{\circ}-2 A=A-18^{\circ}} \\ {108^{\circ}=3 A} \\ {A=36^{\circ}}\end{array}\)

Q.4: \(\text{ If }\tan A=\cot B, \text { prove that } A+B=90^{\circ}\)

Ans : \(\begin{array}{l}{\text { Given that, }} \\ {\tan A=\cot B} \\ {\tan A=\tan \left(90^{\circ}-B\right)} \\ {A=90^{\circ}-B} \\ {A+B=90^{\circ}}\end{array}\)

Q.5: \( \text{ If }\sec 4 A=cosec \left(A-20^{\circ}\right), \text { where } 4 A \text { is an acute angle, find the value of } A.\)

Ans : \(\begin{array}{l}{\text { Given that, }} \\ {\sec 4 A=cosec \left(A-20^{\circ}\right)} \\ {cosec \left(90^{\circ}-4 A\right)=cosec \left(A-20^{\circ}\right)} \\ {90^{\circ}-4 A=A-20^{\circ}} \\ {110^{\circ}=5 A} \\ {A=22^{\circ}}\end{array}\)

Q.6: \(\begin{array}{l}{\text { If } A, \text { Band } C \text { are interior angles of a triangle } A B C \text { then show that }} \\ {\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}}\end{array}\)

Ans : \( \text { We know that for a triangle } A B C \\ \angle A+\angle B+\angle C= 180^{\circ} \\ \angle B+\angle C= 180^{\circ}-\angle A \\ \frac{\angle B+\angle C}{2} =90^{\circ}-\frac{\angle A}{2} \\ \sin \left(\frac{B+C}{2}\right) =\sin \left(90^{\circ}-\frac{A}{2}\right) \\ =\cos \left(\frac{A}{2}\right) \)

Q.7: \(\begin{array}{l}{\text { Express } \sin 67^{\circ}+\cos 75^{\circ} \text { in terms of trigonometric ratios of angles between } 0^{\circ} \text { and }} \\ {45^{\circ} .}\end{array}\)

Ans : \(\begin{array}{l}{\sin 67^{\circ}+\cos 75^{\circ}} \\ {=\sin \left(90^{\circ}-23^{\circ}\right)+\cos \left(90^{\circ}-15^{\circ}\right)} \\ {=\cos 23^{\circ}+\sin 15^{\circ}}\end{array}\)

Q.1: \(\text {Express the trigonometric ratios } \sin A_{\prime} \text { sec } A \text { and tan } A \text { in terms of cot } A.\)

Ans : We know that, \(\begin{array}{l}{cosec ^{2} A=1+\cot ^{2} A} \\ {\frac{1}{cosec ^{2} A}=\frac{1}{1+\cot ^{2} A}} \\ {\sin ^{2} A=\frac{1}{1+\cot ^{2} A}} \\ {\sin A=\pm \frac{1}{\sqrt{1+\cot ^{2} A}}}\end{array}\) \(\sqrt{1+\cot ^{2} A}\) will always be positive as we are adding two positive quantities. Therefore, \(\sin A=\frac{1}{\sqrt{1+\cot ^{2} A}}\) We know that, \(\tan A=\frac{\sin A}{\cos A}\) However, \(\cot A=\frac{\cos A}{\sin A}\) Therefore, \(\tan A=\frac{1}{\cot A}\) Also, \(\sec ^{2} A=1+\tan ^{2} A\) \( =1+\frac{1}{\cot ^{2} A} \\ =\frac{\cot ^{2} A+1}{\cot ^{2} A} \\ \sec A =\frac{\sqrt{\cot ^{2} A+1}}{\cot A} \)

Q.2: \(\text{Write all the other trigonometric ratios of}\angle \mathrm{A} \text { in terms of } \sec A .\)

Ans : We know that, \(\begin{aligned} \cos A &=\frac{1}{\sec A} \\ A|s o,& \sin ^{2} A+\cos ^{2} A=1 \\ \sin ^{2} A &=1-\cos ^{2} A \\ \sin A &=\sqrt{1-\left(\frac{1}{\sec ^{2} A}\right.} )^{2} \\ &=\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}=\frac{\sqrt{\sec ^{2} A-1}}{\sec A} \\ &=\sqrt{\frac{\sec ^{2} A}{\sec ^{2} A}}=\frac{\sqrt{\sec ^{2} A-1}}{\sec A} \\ \tan ^{2} A &=\sec ^{2} A-1 \end{aligned}\) \(\begin{array}{l}{\tan A=\sqrt{\sec ^{2} A-1}} \\ {\cot A=\frac{\cos A}{\sin A}=\frac{\sec A}{\sec A}} \\ {=\frac{1}{\sqrt{\sec ^{2} A-1}}} \\ {cosec A=\frac{1}{\sin A}=\frac{\sec A}{\sqrt{\sec ^{2} A-1}}}\end{array}\)

Q.3: Evaluate 
\(\begin{array}{l}{\text { (i) } \frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}} \\ {\text { (ii) } \sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}}\end{array}\)

Ans : \(\begin{array}{l}{\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}} \\ {=\frac{\left[\sin \left(90^{\circ}-27^{\circ}\right)\right]^{2}+\sin ^{2} 27^{\circ}}{\left[\cos \left(90^{\circ}-73^{\circ}\right)\right]^{2}+\cos ^{2} 73^{\circ}}} \\ {=\frac{\left[\cos 27^{\circ}\right]^{2}+\sin ^{2} 27^{\circ}}{\left[\sin 73^{\circ}\right]^{2}+\cos ^{2} 73^{\circ}}}\end{array}\) \(\begin{array}{l}{=\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\sin ^{2} 73^{\circ}+\cos ^{2} 73^{\circ}}} \\ {=\frac{1}{1}\left(\mathrm{As} \sin ^{2} \mathrm{A}+\cos ^{2} \mathrm{A}=1\right)} \\ {=1}\end{array}\) \(\begin{array}{l}{\text { (ii) } \sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}} \\ {=\left(\sin 25^{\circ}\right)\left\{\cos \left(90^{\circ}-25^{\circ}\right)\right\}+\cos 25^{\circ}\left\{\sin \left(90^{\circ}-25^{\circ}\right)\right\}} \\ {=\left(\sin 25^{\circ}\right)\left(\sin 25^{\circ}\right)+\left(\cos 25^{\circ}\right)\left(\cos 25^{\circ}\right)} \\ {=\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}} \\ {=1\left(\operatorname{As} \sin ^{2} A+\cos ^{2} A=1\right)}\end{array}\)

Q.4: Choose the correct option. Justify your choice.
(i) \(9 \sec ^{2} A-9 \tan ^{2} A=\)
\(\begin{array}{ll}{\text { (A) } 1} & {(\text { B) } 9} & {\text { (C) } 8} & {\text { (D) } 0}\end{array}\)
\(\text {(ii)}(1+\tan \theta+\sec \theta)(1+\cot \theta-cosec \theta)=\)
\(\text { (A) }0 \quad \text { (B) } 1 \text { (C) } 2 \quad \text { (D) }-1\)
\((\text { iii) } \quad(\sec A+\tan A)(1-\sin A)=\)
\(\begin{array}{llll}{\text { (A) } \sec A} & {\text { (B) } \sin A} & {\text { (C) } cosec A} & {\text { (D) } \cos A}\end{array}\)
\((iv) \frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\)
\((A)\sec ^{2} A \quad (B) -1\)
\((\mathrm{C}) \cot ^{2} \mathrm{A} \quad \text { (D) } \tan ^{2} \mathrm{A}\)

Ans : \(\begin{array}{l}{\text { (i) } 9 \mathrm{sec}^{2} \mathrm{A}-9 \tan ^{2} \mathrm{A}} \\ {=9\left(\sec ^{2} \mathrm{A}-\tan ^{2} \mathrm{A}\right)} \\ {=9(1)\left[\operatorname{As} \sec ^{2} \mathrm{A}-\tan ^{2} \mathrm{A}=1\right]} \\ {=9}\end{array}\) Hence, alternatives (B) is correct. (ii) \((1+\tan \theta+\sec \theta)(1+\cot \theta-cosec \theta)\) \(\begin{aligned} &=\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right) \\ &=\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right) \\ &=\frac{(\sin \theta+\cos \theta)^{2}-(1)^{2}}{\sin \theta \cos \theta} \\ &=\frac{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\ &=\frac{2 \sin \theta+\cos \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}=2 \\ &=\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2 \end{aligned}\) Hence, alternative (C) is incorrect. (iii) (secA + tanA)(1-sinA) \(\begin{aligned} &=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A) \\ &=\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A) \\ &=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A} \\ &=\cos A \end{aligned}\) Hence, alternative (D) is correct. (iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}\) \(=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}\) \(=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A\) Hence, alternative (D) is correct.

Q.5: \(\begin{array}{l}{\text { Prove the following identities, where the angles involved are acute angles for which }} \\ {\text { the expressions are defined. }}\end{array}\)

\((i) (cosec \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta} \quad \text { (ii) } \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A\)
\((iii) {\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta cosec \theta} \\ {[\text { Hint } : \text { Write the expression in terms of } \sin \theta \text { and } \cos \theta]}\)
\((iv)\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A} \quad[\text { Hint } : \text { Simplify LHS and RHS separately] }\)
\((v) \frac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosec A+\cot A, \text { using the identity cosec }^{2} A=1+\cot ^{2} A\)
\((v i) \sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A\)
\((vii) \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta\)
\((viii) (\sin A+cosec A)^{2}+(\cos A+\sec A)^{2}=7+\tan ^{2} A+\cot ^{2} A\)
\(\begin{array}{l}{\text { (ix) }(cosec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}} \\ {\text { [Hint : Simplify LHS and RHS separately] }}\end{array}\)
\((x)\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A\)

Ans : \( (i) (cosec \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}\) \(\begin{aligned} \text { L.H.S. } &=(cosec \theta-\cot \theta)^{2} \\ &=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2} \\ &=\frac{(1-\cos \theta)^{2}}{(\sin \theta)^{2}}=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta} \end{aligned}\) \(\begin{aligned} &=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}=\frac{(1-\cos \theta)^{2}}{(1-\cos \theta)(1+\cos \theta)}=\frac{1-\cos \theta}{1+\cos \theta} \\ &=R . H . S_{ .} \end{aligned}\) \((ii) \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A\) \(\begin{aligned} \text { L.H.S. } &=\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A} \\ &=\frac{\cos ^{2} A+(1+\sin A)^{2}}{(1+\sin A)(\cos A)^{2}} \\ &=\frac{\cos ^{2} A+1+\sin ^{2} A+2 \sin A}{(1+\sin A)(\cos A)} \\ &=\frac{\sin ^{2} A+\cos ^{2} A+1+2 \sin A}{(1+\sin A)(\cos A)} \end{aligned}\) \(=\frac{1+1+2 \sin A}{(1+\sin A)(\cos A)}=\frac{2+2 \sin A}{(1+\sin A)(\cos A)}\) \(\begin{aligned} &=\frac{2(1+\sin A)}{(1+\sin A)(\cos A)}=\frac{2}{\cos A}=2 \sec A \\ &=R.H.S. \end{aligned}\) (iii) \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta cosec \theta\) L.H.S.\(=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) \(=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}\) \(=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}+\frac{\cos \theta}{\cos \theta}}\) \(\begin{aligned} &=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)} \\ &=\frac{1}{(\sin \theta-\cos \theta)}\left[\frac{\sin ^{2} \theta}{\cos \theta}-\frac{\cos ^{2} \theta}{\sin \theta}\right] \\ &=\left(\frac{1}{\sin \theta-\cos \theta}\right)\left[\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta}\right] \end{aligned}\) \(\begin{aligned} &=\left(\frac{1}{\sin \theta-\cos \theta}\right)\left[\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta}\right] \\ &=\frac{(1+\sin \theta \cos \theta)}{(\sin \theta \cos \theta)} \end{aligned}\) \( =sec {\theta} cosec {\theta}\\ =R . H . S \) (iv) \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\) \(\mathrm{L.H.S.}=\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{1+\frac{1}{\cos \mathrm{A}}}{\frac{1}{\cos \mathrm{A}}}\) \(\begin{aligned} &=\frac{\cos A+1}{\cos A}=(\cos A+1) \\ &=\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)} \\ &=\frac{1-\cos ^{2} A}{1-\cos A}=\frac{\sin ^{2} A}{1-\cos A} \end{aligned}\) = R.H.S. (v) \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosec A+\cot A\) Using the identity \(cosec ^{2} A=1+\cot ^{2} A\) L.H.S. \(=\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) \(=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}+\frac{1}{\sin A}}\) \(\begin{array}{l}{=\frac{\cot A-1+cosec A}{\cot A+1-cosec A}} \\ {=\frac{\{(\cot A)-(1-cosec A)\}\{(\cot A)-(1-cosec A)\}}{\{(\cot A)+(1-cosec A)\}\{(\cot A)-(1-cosec A)\}}}\end{array}\) \(\begin{aligned} &=\frac{(\cot A-1+cosec A)^{2}}{(\cot A)^{2}-(1-cosec A)^{2}} \\ &=\frac{\cot ^{2} A+1+cosec ^{2} A-2 \cot A-2 cosec A+2 \cot A cosec A}{\cot ^{2} A-\left(1+cosec ^{2} A-2 cosec A\right)} \end{aligned}\) \(\begin{aligned} &=\frac{2 \operatorname{cosec}^{2} A+2 \cot A \cos \epsilon c A-2 \cot A-2 \operatorname{cosec} A}{\cot ^{2} A-1-cosec ^{2} A+2 cosec A} \\ &=\frac{2 cosec A(cosec A+\cot A)-2(\cot A+cosec A)}{\cot ^{2} A-cosec ^{2} A-1+2 cosec A} \end{aligned}\) \(\begin{aligned} &=\frac{(cosec A+\cot A)(2 cosec A-2)}{-1-1+2 cosec A} \\ &=\frac{(cosec A+\cot A)(2 cosec A-2)}{(2 cosec A-2)} \\ &=cosec A+\cot A \\ &=R . H . S \end{aligned}\) (vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A\) \(\begin{aligned} \text { L.H.S. } &=\sqrt{\frac{1+\sin A}{1-\sin A}} \\ &=\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}} \end{aligned}\) \(=\frac{(1+\sin A)}{\sqrt{1-\sin ^{2} A}}=\frac{1+\sin A}{\sqrt{\cos ^{2} A}}\) \(=\frac{1+\sin A}{\cos A} \quad=\sec A+\tan A\) = R.H.S. (vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos \theta \cos \theta}=\tan \theta\) \(\begin{aligned} \text { L.H.S. } &=\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta} \\ &=\frac{\sin \theta\left(1-2 \sin ^{2} \theta\right)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)} \end{aligned}\) \(\begin{aligned} &=\frac{\sin \theta \times\left(1-2 \sin ^{2} \theta\right)}{\cos \theta \times\left\{2\left(1-\sin ^{2} \theta\right)-1\right\}} \\ &=\frac{\sin \theta \times\left(1-2 \sin ^{2} \theta\right)}{\cos \theta \times\left(1-2 \sin ^{2} \theta\right)} \\ &=\tan \theta=R .H .S \end{aligned}\) \((viii)(\sin A+cosec A)^{2}+(\cos A+\sec A)^{2}=7+\tan ^{2} A+\cot ^{2} A\) L.H.S\(=(\sin A+cosec A)^{2}+(\cos A+\sec A)^{2}\) \(\begin{array}{l}{=\sin ^{2} A+\operatorname{cosec}^{2} A+2 \sin A cosec A+\cos ^{2} A+\sec ^{2} A+2 \cos A \sec A} \\ {=\left(\sin ^{2} A+\cos ^{2} A\right)+\left(cosec ^{2} A+\sec ^{2} A\right)+2 \sin A\left(\frac{1}{\sin A}\right)+2 \cos A\left(\frac{1}{\cos A}\right)} \\ {=(1)+\left(1+\cot ^{2} A+1+\tan ^{2} A\right)+(2)+(2)} \\ {=7+\tan ^{2} A+\cot ^{2} A} \\ {=R . H . S}\end{array}\) \((ix){(cosec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}} \\ {\text { L.H.S }=(cosec A-\sin A)(\sec A-\cos A)}\) \(\begin{aligned} &=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) \\ &=\left(\frac{1-\sin ^{2} A}{\sin A}\right)\left(\frac{1-\cos ^{2} A}{\cos A}\right) \\ &=\frac{\left(\cos ^{2} A\right)\left(\sin ^{2} A\right)}{\sin A \cos A} \\ &=\sin A \cos A \end{aligned}\) R.H.S\(=\frac{1}{\tan A+\cot A}\) \(\begin{aligned} &=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\frac{1}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}} \\ &=\frac{\sin A \cos A}{\sin ^{2} A+\cos ^{2} A}=\sin A \cos A \end{aligned}\) Hence, L.H.S = R.H.S \((x)\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A\) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A}{\sin ^{2} A}}\) \(\begin{aligned}=& \frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}=\frac{\sin ^{2} A}{\cos ^{2} A} \\=& \tan ^{2} A \end{aligned}\) \(\begin{aligned}\left(\frac{1-\tan A}{1-\cot A}\right)^{2} &=\frac{1+\tan ^{2} A-2 \tan A}{1+\cot ^{2} A-2 \cot A} \\ &=\frac{\sec ^{2} A-2 \tan A}{\cos e c^{2} A-2 \cot A} \end{aligned}\) \(=\frac{\frac{1}{\cos ^{2} A}-\frac{2 \sin A}{\cos A}}{\frac{1}{\sin ^{2} A}-\frac{2 \cos A}{\sin A}}=\frac{\frac{1-2 \sin A \cos A}{\cos ^{2} A}}{\frac{1-2 \sin ^{2} A}{\sin ^{2} A}}\) \(=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A\)

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