NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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) 
\(\begin{array}{l}{ \sin \mathrm{C}=\frac{\text { Side opposite to } \angle \mathrm{C}}{\text { Hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}} \\ {=\frac{24}{25}}\end{array}\)
\(\begin{array}{l}{\cos \mathrm{C}=\frac{\text { Side adjacent to } \angle \mathrm{C}}{\mathrm{Hypotenuse}}=\frac{\mathrm{BC}}{\mathrm{AC}}} \\ {=\frac{7}{25}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { Applying Pythagoras theorem for } \Delta P Q R, \text { we obtain }} \\ {P R^{2}=P Q^{2}+Q R^{2}} \\ {(13 \mathrm{cm})^{2}=(12 \mathrm{cm})^{2}+Q R^{2}} \\ {169 \mathrm{cm}^{2}=144 \mathrm{cm}^{2}+Q R^{2}} \\ {25 \mathrm{cm}^{2}=Q R^{2}} \\ {Q R=5 \mathrm{cm}}\end{array}\)

\(\begin{aligned} \tan \mathbf{P} &=\frac{\text { Side opposite to } \angle P}{\text { Side adjacent to } \angle P}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \\ \cot R &=\frac{\text { Side adjacent to } \angle R}{\text { Side opposite to } \angle R}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \end{aligned}\)
\(\tan \mathrm{P}-\cot \mathrm{R}=\frac{5}{12}-\frac{5}{12}=0\) 

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
\(\begin{array}{l}{A C^{2}=A B^{2}+B C^{2}} \\ {(4 k)^{2}=A B^{2}+(3 k)^{2}} \\ {16 k^{2}-9 k^{2}=A B^{2}} \\ {7 k^{2}=A B^{2}} \\ {A B=\sqrt{7} k}\end{array}\)
\(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotent } t 0 \angle A} \\ &=\frac{A B}{A C}=\frac{\sqrt{7 k}}{4 k}=\frac{\sqrt{7}}{4} \\ \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A} \\ &=\frac{B C}{A B}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}} \end{aligned}\) 

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
\(\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}} \\ {=(8 k)^{2}+(15 k)^{2}} \\ {=64 k^{2}+225 \mathrm{k}^{2}} \\ {=289 \mathrm{k}^{2}} \\ {\mathrm{AC}=17 k}\end{array}\)
\(\begin{aligned} \sin A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C} \\ &=\frac{15 k}{17 k}=\frac{15}{17} \\ \sec A &=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A} \\ &=\frac{A C}{A B}=\frac{17}{8} \end{aligned}\) 

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
\(\sin \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}\)
\(\cos \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}\)
\(\tan \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Side adjacent to } \angle \theta}=\frac{B C}{A B}=\frac{5 k}{12 k}=\frac{5}{12}\)
\(\cot \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Side opposite to } \angle \theta}=\frac{A B}{B C}=\frac{12 k}{5 k}=\frac{12}{5}\)
\(cosec \ \theta=\frac{\text { Hypotenuse }}{\text { Side opposite to } \angle \theta}=\frac{A C}{B C}=\frac{13 k}{5 k}=\frac{13}{5}\) 

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) 

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) \(\cot ^{2} \theta=(\cot \theta)^{2}=\left(\frac{7}{8}\right)^{2}=\frac{49}{64}\) 

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 \(\triangle \mathrm{ABC}\),
\(\begin{array}{l}{(A C)^{2}=(A B)^{2}+(B C)^{2}} \\ {=(4 k)^{2}+(3 k)^{2}} \\ {=16 k^{2}+9 k^{2}} \\ {=25 k^{2}} \\ {A C=5 k}\end{array}\)
\(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }}=\frac{A B}{A C} \\ &=\frac{4 k}{5 k}=\frac{4}{5} \end{aligned}\)
\(\begin{aligned} \sin A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C} \\ &=\frac{3 k}{5 k}=\frac{3}{5} \end{aligned}\)
\(\begin{aligned} \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A B} \\ &=\frac{3 k}{4 k}=\frac{3}{4} \end{aligned}\)
\(\begin{array}{c}{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}} \\ {=\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}} \end{array}\)
\(\begin{array}{l}{\cos ^{2} A-\sin ^{2} A=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}} \\ {=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}} \\ {\therefore \quad \frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A}\end{array}\) 

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 
\(=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0\) 

Ans : Given that, PR + QR = 25
PQ = 5
Let PR be x. 
Therefore, QR = 25 -x

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

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
\(\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A}=\frac{12}{5}\)
\(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{12}{5}\)
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  

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) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
\(\begin{aligned} &=\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\=& \frac{5\left(\frac{1}{4}\right)+\left(\frac{16}{3}\right)-1}{\frac{1}{4}+\frac{3}{4}} \\=& \frac{15+64-12}{\frac{4}{4}}=\frac{67}{12} \end{aligned}\) 

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. 

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

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} 

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

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

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

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

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

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

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

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, \(\sec ^{2} A=1+\tan ^{2} A\)
\( =1+\frac{1}{\cot ^{2} A} \\ =\frac{\cot ^{2} A+1}{\cot ^{2} A} \\ \sec A =\frac{\sqrt{\cot ^{2} A+1}}{\cot A} \) 

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

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

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. 

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
\((x)\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A\)
\(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A}{\sin ^{2} A}}\)
\(\begin{aligned}=& \frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}=\frac{\sin ^{2} A}{\cos ^{2} A} \\=& \tan ^{2} A \end{aligned}\)
\(\begin{aligned}\left(\frac{1-\tan A}{1-\cot A}\right)^{2} &=\frac{1+\tan ^{2} A-2 \tan A}{1+\cot ^{2} A-2 \cot A} \\ &=\frac{\sec ^{2} A-2 \tan A}{\cos e c^{2} A-2 \cot A} \end{aligned}\)
\(=\frac{\frac{1}{\cos ^{2} A}-\frac{2 \sin A}{\cos A}}{\frac{1}{\sin ^{2} A}-\frac{2 \cos A}{\sin A}}=\frac{\frac{1-2 \sin A \cos A}{\cos ^{2} A}}{\frac{1-2 \sin ^{2} A}{\sin ^{2} A}}\)
\(=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A\)