Question 14 Marks
$
\text { Given } \cos 38^{\circ} \sec \left(90^{\circ}-2 A\right)=1 \text {, Find the value of } \angle \mathrm{A}
$
Answer$
\begin{aligned}
& \cos 38^{\circ} \sec \left(90^{\circ}-2 A\right)=1 \\
& \Rightarrow \cos 38^{\circ} \cos e c 2 A=1 \\
& \Rightarrow \cos 38^{\circ}\left(\frac{1}{\sin 2 A}\right)=1 \\
& \Rightarrow \sin 2 A=\cos \left(90-52^{\circ}\right) \\
& \Rightarrow \sin 2 A=\sin 52^{\circ} \\
& \Rightarrow 2 A=52^{\circ} \\
& \Rightarrow A=26^{\circ}
\end{aligned}
$
View full question & answer→Question 24 Marks
$
\text { Find } x \text {, if } \cos (2 x-6)=\cos ^2 30^{\circ}-\cos ^2 60^{\circ}
$
Answer$
\begin{aligned}
& \cos (2 x-6)=\cos ^2 30^{\circ}-\cos ^2 60^{\circ} \\
& \Rightarrow \cos (2 x-6)=\cos ^2\left(90^{\circ}-60^{\circ}\right)-\cos ^2 60^{\circ} \\
& \Rightarrow \cos (2 x-6)=\sin ^2 60^{\circ}-\cos ^2 60^{\circ} \\
& \Rightarrow \cos (2 x-6)=1-2 \cos ^2 60^{\circ}=1-2\left(\frac{1}{2}\right)^2=1-\frac{1}{2}=\frac{1}{2} \\
& \Rightarrow \cos (2 x-6)=\frac{1}{2} \\
& \Rightarrow \cos (2 x-6)=\cos 60^{\circ} \\
& \Rightarrow(2 x-6)=60^{\circ} \\
& \Rightarrow 2 x=66^{\circ} \\
& \Rightarrow x=33^{\circ}
\end{aligned}
$
View full question & answer→Question 34 Marks
$
\text { Find the value of } \mathrm{x} \text {, if } \cos x=\cos 60^{\circ} \cos 30^{\circ}-\sin 60^{\circ} \sin 30^{\circ}
$
Answer$
\begin{aligned}
& \cos (2 x-6)=\cos ^2 30^{\circ}-\cos ^2 60^{\circ} \\
& \Rightarrow \cos (2 x-6)=\cos ^2\left(90^{\circ}-60^{\circ}\right)-\cos ^2 60^{\circ} \\
& \Rightarrow \cos (2 x-6)=\sin ^2 60^{\circ}-\cos ^2 60^{\circ} \\
& \Rightarrow \cos (2 x-6)=1-2 \cos ^2 60^{\circ}=1-2\left(\frac{1}{2}\right)^2=1-\frac{1}{2}=\frac{1}{2} \\
& \Rightarrow \cos (2 x-6)=\frac{1}{2} \\
& \Rightarrow \cos (2 x-6)=\cos 60^{\circ} \\
& \Rightarrow(2 x-6)=60^{\circ} \\
& \Rightarrow 2 x=66^{\circ} \\
& \Rightarrow x=33^{\circ}
\end{aligned}
$
View full question & answer→Question 44 Marks
$
\text { If } \tan A+\sin A=m \text { and } \tan A-\sin A=n \text {, prove that }\left(m^2-n^2\right)^2=16 m n
$
Answer$
\begin{aligned}
& \text { Consider }\left(m^2-n^2\right)=(\tan A+\sin A)^2-(\tan A-\sin A)^2 \\
& \Rightarrow[\tan \mathrm{A}+\sin \mathrm{A}-(\tan \mathrm{A}-\sin \mathrm{A})][\tan \mathrm{A}+\sin \mathrm{A}+(\tan \mathrm{A}-\sin \mathrm{A})] \\
& \Rightarrow[2 \sin \mathrm{A}][2 \tan \mathrm{A}]=4 \sin \mathrm{A} \tan \mathrm{A}
\end{aligned}
$
Now LHS $=\left(m^2-n^2\right)^2=(4 \sin A \tan A)^2=16 \sin ^2 A \tan ^2 A$
Also, RHS $=16 m n=16(\tan A+\sin A)(\tan A-\sin A)$
$
\begin{aligned}
& \Rightarrow \text { RHS }=16 \mathrm{mn}=16\left(\tan ^2 A-\sin ^2 A\right)=16\left(\frac{\sin ^2 A}{\cos ^2 A}-\sin ^2 A\right) \\
& \Rightarrow 16 \sin ^2 A\left(\frac{1-\cos ^2 A}{\cos ^2 A}\right)=16 \sin ^2 A\left(\frac{\sin ^2 A}{\cos ^2 A}\right)=16 \sin ^2 A \tan ^2 A
\end{aligned}
$
Thus, $\left(m^2-n^2\right)^2=16 m n$
View full question & answer→Question 54 Marks
$
\text { If } \frac{x}{a \cos \theta}=\frac{y}{b \sin \theta} \text { and } \frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2, \text { prove that } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$
AnswerLet $\frac{x}{a \cos \theta}=\frac{y}{b \sin \theta}$$\ldots(1)$
and $\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2$$\ldots(2)$
From (1), $\frac{y}{\sin \theta}=\frac{x b}{a \cos \theta}$
Putting (2), we get $\frac{a x}{\cos \theta}-b \frac{x b}{a \cos \theta}=a^2-b^2$
$
\begin{aligned}
& \Rightarrow \frac{a x}{\cos \theta}-\frac{x b^2}{a \cos \theta}=a^2-b^2 \\
& \Rightarrow \frac{x\left(a^2-b^2\right)}{a \cos \theta}=a^2-b^2
\end{aligned}
$
$
\Rightarrow \mathrm{x}=\mathrm{a} \cos \theta
$
$
\text { By(1), } y=\frac{x b \sin \theta}{a \cos \theta}=\frac{a \cos \theta b \sin \theta}{a \cos \theta}=b \sin \theta
$
Thus, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{a^2 \cos ^2 \theta}{a^2}+\frac{b^2 \sin ^2 \theta}{b^2}=\sin ^2 \theta+\cos ^2 \theta=1$
View full question & answer→Question 64 Marks
Prove the following identity :
$
\frac{\sec ^2 \theta-\sin ^2 \theta}{\tan ^2 \theta}=\operatorname{cosec}{ }^2 \theta-\cos ^2 \theta
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\sec ^2 \theta-\sin ^2 \theta}{\tan ^2 \theta} \\
& =\frac{\frac{1}{\cos ^2 \theta}-\sin ^2 \theta}{\frac{\sin ^2 \theta}{\cos ^2 \theta}} \\
& =\frac{1-\sin ^2 \theta \cos ^2 \theta}{\frac{\cos ^2 \theta}{\frac{\sin ^2 \theta}{\cos ^2 \theta}}} \\
& =\frac{1-\sin ^2 \theta \cos ^2 \theta}{\sin ^2 \theta} \\
& =\frac{1}{\sin ^2 \theta}-\frac{\sin ^2 \theta \cos ^2 \theta}{\sin ^2 \theta} \\
& =\operatorname{cosec} \cos ^2 \theta-\cos ^2 \theta
\end{aligned}
$
View full question & answer→Question 74 Marks
Prove the following identity:
$
\frac{\tan ^2 \theta}{\tan ^2 \theta-1}+\frac{\operatorname{cosec}{ }^2 \theta}{\sec ^2 \theta-\operatorname{cosec}^2 \theta}=\frac{1}{\sin ^2 \theta-\cos ^2 \theta}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\tan ^2 \theta}{\tan ^2 \theta-1}+\frac{\operatorname{cosec} 2}{\sec ^2 \theta-\operatorname{cosec}{ }^2 \theta} \\
& =\frac{\frac{\sin ^2 \theta}{\cos ^2 \theta}}{\frac{\sin ^2 \theta}{\cos ^2 \theta}-1}+\frac{\frac{1}{\sin ^2 \theta}}{\frac{1}{\cos ^2 \theta}-\frac{1}{\sin ^2 \theta}} \\
& =\frac{\sin ^2 \theta}{\sin ^2 \theta-\cos ^2 \theta}+\frac{\frac{1}{\sin ^2 \theta}}{\frac{\sin ^2 \theta-\cos ^2 \theta}{\cos ^2 \theta \sin ^2 \theta}} \\
& =\frac{\sin ^2 \theta}{\sin ^2 \theta-\cos ^2 \theta}+\frac{\cos ^2 \theta}{\sin ^2 \theta-\cos ^2 \theta} \\
& =\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta-\cos ^2 \theta}=\frac{1}{\sin ^2 \theta-\cos ^2 \theta} \quad\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)
\end{aligned}
$
View full question & answer→Question 84 Marks
Prove the following identity:
$
\sin ^8 \theta-\cos ^8 \theta=\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)
$
Answer$
\begin{aligned}
& \text { LHS }=\sin ^8 \theta-\cos ^8 \theta \\
& =\left(\sin ^4 \theta\right)^2-\left(\cos ^4 \theta\right)^2 \\
& =\left(\sin ^4 \theta-\cos ^4 \theta\right)\left(\sin ^4 \theta+\cos ^4 \theta\right) \\
& =\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(\sin ^2 \theta+\cos ^2 \theta\right)\left(\sin ^4 \theta+\cos ^4 \theta\right) \\
& =\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(\sin ^4 \theta+\cos ^4 \theta\right) \\
& =\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2+2 \sin ^2 \theta \cos ^2 \theta-2 \sin ^2 \theta \cos ^2 \theta\right) \\
& =\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta\right) \\
& =\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)
\end{aligned}
$
View full question & answer→Question 94 Marks
Prove the following identity:
$
2\left(\sin ^6 \theta+\cos ^6 \theta\right)-3\left(\sin ^4 \theta+\cos ^4 \theta\right)+1=0
$
Answer$
\begin{aligned}
& \text { LHS }=2\left(\sin ^6 \theta+\cos ^6 \theta\right)-3\left(\sin ^4 \theta+\cos ^4 \theta\right)+1 \\
& =2\left(\sin ^6 \theta+\cos ^6 \theta\right)-3\left(\sin ^4 \theta+\cos ^4 \theta\right)+1 \\
& =2\left[\left(\sin ^2 \theta\right)^3+\left(\cos ^2 \theta\right)^3\right]-3\left(\sin ^4 \theta+\cos ^4 \theta\right)+1 \\
& =2\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)\left\{\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2-\sin ^2 \theta \cos ^2 \theta\right\}\right]-3\left(\sin ^4 \theta+\cos ^4 \theta\right)+1 \\
& =2\left\{\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2-\sin ^2 \theta \cos ^2 \theta\right\}-3\left(\sin ^4 \theta+\cos ^4 \theta\right)+1 \\
& =2 \sin ^4 \theta+2 \cos ^4 \theta-2 \sin ^2 \theta \cos ^2 \theta-3 \sin ^4 \theta-3 \cos ^4 \theta+1 \\
& =-\sin ^4 \theta-\cos ^4 \theta-2 \sin ^2 \theta \cos ^2 \theta+1 \\
& =-\left(\sin ^4 \theta+\cos ^4 \theta+2 \sin ^2 \theta \cos ^2 \theta\right)+1 \\
& =-\left(\sin ^2 \theta+\cos ^2 \theta\right)^2+1=-1+1=0
\end{aligned}
$
View full question & answer→Question 104 Marks
Prove the following identity:
$
(1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\operatorname{cosec} 2}-\frac{\operatorname{cosec} A}{\sec ^2 A}
$
Answer$
\begin{aligned}
& \text { LHS }=(1+\cot A+\tan A)(\sin A-\cos A) \\
& =\left(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)(\sin A-\cos A) \\
& =\left(\frac{\sin A \cos A+\cos ^2 A+\sin ^2 A}{\sin A \cos A}\right)(\sin A-\cos A) \\
& =\frac{\left(\sin ^3 A-\cos ^3 A\right)}{\sin A \cos A}\left(\because\left(\sin ^3 A-\cos ^3 A\right)=(\sin A-\cos A)\left(\sin A \cos A+\cos ^2 A+\sin ^2 A\right)\right)
\end{aligned}
$
$
\begin{aligned}
& =\frac{\sin ^3 A}{\sin A \cos A}-\frac{\cos ^3 A}{\sin A \cos A} \\
& =\frac{\sin ^2 A}{\cos A}-\frac{\cos ^2 A}{\sin A}=\frac{1}{\cos A} \times \sin ^2 A-\frac{1}{\sin A} \times \cos ^2 A \\
& =\sec A \sin ^2 A-\operatorname{cosec} A \cos ^2 A \\
& =\frac{\sec A}{\operatorname{cosec} 2}-\frac{\operatorname{cosec} A}{\sec ^2 A}
\end{aligned}
$
View full question & answer→Question 114 Marks
Prove the following identity:
$
\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}=\frac{\cos A+1}{\sin A}
$
Answer$
\begin{aligned}
& \frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1} \\
& =\frac{\cot A+\operatorname{cosec} A-\left(\operatorname{cosec}{ }^2 A-\cot ^2 A\right)}{\cot A-\operatorname{cosec} A+1}\left[\operatorname{cosec}^2 A-\cot ^2 A=1\right] \\
& =\frac{\cot A+\operatorname{cosec} A-[(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)]}{\cot A-\operatorname{cosec} A+1} \\
& =\frac{\cot A+\operatorname{cosec} A[1-\operatorname{cosec} A+\cot A]}{\cot A-\operatorname{cosec} A+1} \\
& =\cot A+\operatorname{cosec} A \\
& =\frac{\cos A}{\sin A}+\frac{1}{\sin A} \\
& =\frac{1+\cos A}{\sin A}
\end{aligned}
$
View full question & answer→Question 124 Marks
Prove the following identity:
$
\frac{1}{\cos A+\sin A-1}+\frac{2}{\cos A+\sin A+1}=\operatorname{cosec} A+\sec A
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{1}{(\cos A+\sin A)-1}+\frac{1}{(\cos A+\sin A)+1} \\
& =\frac{\cos A+\sin A+1+\cos A+\sin A-1}{(\cos A+\sin A)^2-1} \\
& =\frac{2(\cos A+\sin A)}{\cos ^2 A+\sin ^2 A+2 \cos A \sin A-1} \\
& =\frac{2(\cos A+\sin A)}{1+2 \cos A \sin A-1}=\frac{\cos A+\sin A}{\cos A \sin A} \\
& =\frac{\cos A}{\cos A \sin A}+\frac{\sin A}{\cos A \sin A} \\
& =\frac{1}{\sin A}+\frac{1}{\cos A} \\
& =\operatorname{cosec} A+\sec A
\end{aligned}
$
View full question & answer→Question 134 Marks
Prove the following identity:
$
\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\sin A+\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A-\sin B} \\
& =\frac{(\sin A+\sin B)(\sin A-\sin B)+(\cos A+\cos B)(\cos A-\cos B)}{(\cos A+\cos B)(\sin A-\sin B)} \\
& =\frac{\sin ^2 A-\sin ^2 B+\cos ^2 A-\cos ^2 B}{(\cos A+\cos B)(\sin A-\sin B)} \\
& =\frac{\left(\sin ^2 A+\cos ^2 A\right)-\left(\sin ^2 B+\cos ^2 B\right)}{(\cos A+\cos B)(\sin A-\sin B)} \\
& =\frac{1-1}{(\cos A+\cos B)(\sin A-\sin B)} \\
& =\frac{0}{(\cos A+\cos B)(\sin A-\sin B)} \\
& =0 \\
& \frac{\sin A+\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A-\sin B}=0
\end{aligned}
$
View full question & answer→Question 144 Marks
Prove the following identity:
$
\left(\tan \theta+\frac{1}{\cos \theta}\right)^2+\left(\tan \theta-\frac{1}{\cos \theta}\right)^2=2\left(\frac{1+\sin ^2 \theta}{1-\sin ^2 \theta}\right)
$
Answer$
\begin{aligned}
& \left(\tan \theta+\frac{1}{\cos \theta}\right)^2+\left(\tan \theta-\frac{1}{\cos \theta}\right)^2 \\
& =\left(\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)^2+\left(\frac{\sin \theta}{\cos \theta}-\frac{1}{\cos \theta}\right)^2 \\
& =\left(\frac{\sin \theta+1}{\cos \theta}\right)^2+\left(\frac{\sin \theta-1}{\cos \theta}\right)^2 \\
& =\frac{(\sin \theta+1)^2}{\cos ^2 \theta}+\frac{(\sin \theta-1)^2}{\cos ^2 \theta} \\
& =\frac{(\sin \theta+1)^2+(\sin \theta-1)^2}{\cos ^2 A} \\
& =\frac{\sin { }^2 \theta+1+2 \sin \theta+\sin ^2 \theta+1-2 \sin \theta}{1-\sin ^2 \theta} \\
& =\frac{2\left(1+\sin ^2 \theta\right)}{1-\sin ^2 \theta}
\end{aligned}
$
View full question & answer→Question 154 Marks
Prove the following identity:
$
\frac{\cos A}{1-\tan A}+\frac{\sin ^2 A}{\sin A-\cos A}=\cos A+\sin A
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\cos A}{1-\tan A}+\frac{\sin ^2 A}{\sin A-\cos A} \\
& =\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin ^2 A}{\sin A-\cos A} \\
& =\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin ^2 A}{\sin A-\cos A} \\
& =\frac{\cos ^2 A}{(\cos A-\sin A)}-\frac{\sin ^2 A}{(\cos A-\sin A)} \\
& =\frac{\cos ^2 A-\sin 2}{\cos A-\sin A}=\frac{(\cos A+\sin A)(\cos A-\sin A)}{(\cos A-\sin A)} \\
& =(\cos A+\sin A)
\end{aligned}
$
View full question & answer→Question 164 Marks
Prove the following identity:
$
\tan ^2 A-\tan ^2 B=\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cos ^2 B}
$
Answer$
\begin{aligned}
& \text { LHS }=\tan ^2 A-\tan ^2 B \\
& =\frac{\sin ^2 A}{\cos ^2 A}-\frac{\sin ^2 B}{\cos ^2 B} \\
& =\frac{\sin ^2 A \cos ^2 B-\cos ^2 A \sin ^2 B}{\cos ^2 A \cos ^2 B} \\
& =\frac{\left(1-\cos ^2 A\right) \cos ^2 B-\cos ^2 A\left(1-\cos ^2 B\right)}{\cos ^2 A \cos ^2 B} \\
& =\frac{\cos ^2 B-\cos ^2 A \cos ^2 B-\cos ^2 A+\cos ^2 A \cos ^2 B}{\cos ^2 A \cos ^2 B} \\
& =\frac{\cos ^2 B-\cos ^2 A}{\cos ^2 A \cos ^2 B} \\
& =\frac{\left(1-\sin ^2 B\right)-\left(1-\sin ^2 A\right)}{\cos ^2 A \cos ^2 B} \\
& =\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cos ^2 B}
\end{aligned}
$
Hence $\tan ^2 A-\tan ^2 B=\frac{\cos ^2 B-\cos ^2 A}{\cos ^2 A \cos ^2 B}=\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cos ^2 B}$
View full question & answer→Question 174 Marks
Prove the following identity:
$
\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2 \sin ^2 A-1}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A} \\
& =\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A+\cos A)(\sin A-\cos A)} \\
& =\frac{\sin ^2 A+\cos ^2 A+2 \sin A \cos A+\sin ^2 A+\cos ^2 A-2 \sin A \cos A}{\sin ^2 A-\cos ^2 A} \\
& =\frac{2\left(\sin ^2 A+\cos ^2 A\right)}{\sin ^2 A-\cos ^2 A} \\
& =\frac{2}{\sin ^2 A-\cos ^2 A}\left[\sin ^2 A+\cos ^2 A=1\right] \\
& =\frac{2}{\sin ^2 A-\cos ^2 A}=\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)} \\
& \Rightarrow \frac{2}{2 \sin ^2 A-1}
\end{aligned}
$
View full question & answer→Question 184 Marks
Prove the following identity:
$
\sqrt{\frac{\sec q-1}{\sec q+1}}+\sqrt{\frac{\sec q+1}{\sec q-1}}=2 \operatorname{coses} q
$
Answer$
\begin{aligned}
& \sqrt{\frac{\sec q-1}{\sec q+1}}+\sqrt{\frac{\sec q+1}{\sec q-1}} \\
& =\sqrt{\frac{\sec q-1}{\sec q+1} \cdot \frac{\sec q-1}{\sec q-1}}+\sqrt{\frac{\sec q+1}{\sec q-1} \cdot \frac{\sec q+1}{\sec q+1}} \\
& =\sqrt{\frac{(\sec q-1)^2}{\sec ^2 q-1}}+\sqrt{\frac{(\sec q+1)^2}{\sec q^2-1}} \\
& =\sqrt{\frac{(\sec q-1)^2}{\tan ^2 q}}+\sqrt{\frac{(\sec q+1)^2}{\tan ^2 q}}\left(Q \sec ^2 q-1=\tan ^2 q\right) \\
& =\frac{\sec q-1}{\tan q}+\frac{\sec q+1}{\tan q}=\frac{\sec q-1+\sec q+1}{\tan q} \\
& =\frac{2 \sec q}{\tan q}=\frac{\frac{2}{\cos q}}{\frac{\sin q}{\cos q}}=\frac{2}{\sin q}=2 \cos e c q \\
&
\end{aligned}
$
View full question & answer→Question 194 Marks
Prove the following identity:
$
\sqrt{\frac{1+\sin q}{1-\sin q}}+\sqrt{\frac{1-\sin q}{1+\sin q}}=2 \text { secq }
$
Answer$
\begin{aligned}
& \sqrt{\frac{1+\sin q}{1-\sin q}}+\sqrt{\frac{1-\sin q}{1+\sin q}} \\
& =\sqrt{\frac{1+\sin q}{1-\sin q} \cdot \frac{1+\sin q}{1+\sin q}}+\sqrt{\frac{1-\sin q}{1+\sin q} \cdot \frac{1-\sin q}{1-\sin q}} \\
& =\sqrt{\frac{(1+\sin q)^2}{1-\sin ^2 q}}+\sqrt{\frac{(1-\sin q)^2}{1-\sin ^2 q}}=\sqrt{\frac{(1+\sin q)^2}{\cos ^2 q}}+\sqrt{\frac{(1-\sin q)^2}{\cos ^2 q}} \\
& =\frac{1+\sin q}{\cos q}+\frac{1-\sin q}{\cos q}=\frac{1+\sin q+1-\sin q}{\cos q}=\frac{2}{\cos q} \\
& =2 \sec q
\end{aligned}
$
View full question & answer→Question 204 Marks
Prove the following identity:
$
(\cot A-\operatorname{cosec} A)^2=\frac{1-\cos A}{1+\cos A}
$
Answer$
\begin{aligned}
& \text { LHS }=(\cot A-\cos e c A)^2 \\
& =\left[\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right]^2 \\
& =\left[\frac{\cos A-1}{\sin A}\right]^2 \\
& =\frac{(\cos A-1)^2}{\sin ^2 A}=\frac{(\cos A-1)^2}{1-\cos ^2 A} \\
& =\frac{(-(1-\cos A))^2}{(1-\cos A)(1+\cos A)}=\frac{(1-\cos A)(1-\cos A)}{(1-\cos A)(1+\cos A)} \\
& =\frac{1-\cos A}{1+\cos A}
\end{aligned}
$
View full question & answer→Question 214 Marks
Prove the following identities:
$
\frac{\sec A-1}{\sec A+1}=\left(\frac{\sin A}{1+\cos A}\right)^2
$
Answer$
\begin{aligned}
& \frac{\sec \mathrm{A}-1}{\sec \mathrm{A}+1}=\left(\frac{\sin \mathrm{A}}{1+\cos \mathrm{A}}\right)^2 \\
& =\text { L.H.S. } \\
& =\frac{\sec \mathrm{A}-1}{\sec \mathrm{A}+1} \\
& =\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1} \\
& =\frac{1-\cos A}{1+\cos A}
\end{aligned}
$
Multiplied by $1+\cos A$
$
\begin{aligned}
& =\frac{1-\cos ^2 A}{(1+\cos A)^2} \\
& =\frac{\sin ^2 A}{(1+\cos A)^2} \\
& =\left(\frac{\sin A}{1+\cos A}\right)^2 \\
& =\text { R.H.S }
\end{aligned}
$
Hence Proved.
View full question & answer→Question 224 Marks
Prove the following identities:
$
\frac{\tan A+\tan B}{\cot A+\cot B}=\tan A \tan B
$
Answer$
\begin{aligned}
& \frac{\tan A+\tan B}{\cot A+\cot B}=\tan A \tan B \\
& =\text { L.H.S. } \\
& =\frac{\tan A+\tan B}{\cot A+\cot B} \\
& =\frac{\tan A+\tan B}{\frac{1}{\tan A}+\frac{1}{\tan B}} \\
& =\frac{\tan A+\tan B}{\frac{\tan A+\tan B}{\tan A \cdot \tan B}} \\
& =\frac{(\tan A+\tan B)(\tan A \cdot \tan B)}{\tan A+\tan B} \\
& =\tan A \tan B \\
& =\text { R.H.S. }
\end{aligned}
$
Hence, proved.
View full question & answer→Question 234 Marks
Prove the following identity :
$
(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}
$
Answer$
\begin{aligned}
& \text { LHS }=(\operatorname{cosec} A-\sin A)(\sec A-\cos A) \\
& =\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) \\
& =\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)=\cos A \cdot \sin A \\
& \text { RHS }=\frac{1}{\tan A+\cot A} \\
& =\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\frac{1}{\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 \cdot \cos ^2}}=\cos A \cdot \sin A
\end{aligned}
$
Hence, $\mathrm{LHS}=\mathrm{RHS}$
View full question & answer→Question 244 Marks
Prove the following identity :
$
\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}=\frac{\cos A+1}{\sin A}
$
Answer$
\begin{aligned}
& \frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}=\frac{\cos A+1}{\sin A} \\
& \text { LHS }=\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}
\end{aligned}
$
we know that, $\operatorname{cosec}^2 \mathrm{~A}-\cot ^2 \mathrm{~A}=1$
substituting this in the numerator
$
\begin{aligned}
& \frac{\operatorname{cosec} A+\cot A-\left(\operatorname{cosec}{ }^2 A-\cot ^2 A\right)}{\cot A-\operatorname{cosec} A+1} \ldots . . .\left(x^2-\mathrm{y}^2=(\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})\right) \\
& \frac{\operatorname{cosec} A+\cot A-(\operatorname{cosec} A+\cot A)(\operatorname{cosec} A-\cot A)}{\cot A-\operatorname{cosec} A+1}
\end{aligned}
$
taking common
$
\frac{(\operatorname{cosec} A+\cot A)(1-\operatorname{cosec} A+\cot A)}{\cot A-\operatorname{cosec} A+1}
$
cancelling like terms in numerator and denominator we are left with $\operatorname{cosec} A+\cot A$
$
\begin{aligned}
& =\frac{1}{\sin A}+\frac{\cos A}{\sin A} \\
& =\frac{1+\cos A}{\sin A} \\
& =\text { RHSz }
\end{aligned}
$
View full question & answer→Question 254 Marks
Prove the following identity:
$
\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{1+\sin \theta}{\cos \theta}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1} \\
& =\frac{\tan \theta+\sec \theta-\left\{\sec ^2 \theta-\tan ^2 \theta\right\}}{1+\tan \theta-\sec \theta} \\
& =\frac{\tan \theta+\sec \theta-\{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)\}}{1+\tan \theta-\sec \theta} \\
& =\frac{[\tan \theta+\sec \theta]\{1-(\sec \theta-\tan \theta)\}}{[1+\tan \theta-\sec \theta]}=\frac{[\tan \theta+\sec \theta][1+\tan \theta-\sec \theta]}{[1+\tan \theta-\sec \theta]} \\
& =[\tan \theta+\sec \theta]=\frac{1+\sin \theta}{\cos \theta}=\text { RHS }
\end{aligned}
$
View full question & answer→Question 264 Marks
Prove the following identity :
$
\frac{\cot ^2 \theta(\sec \theta-1)}{(1+\sin \theta)}=\sec ^2 \theta\left(\frac{1-\sin \theta}{1+\sec \theta}\right)
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\cot ^2 \theta(\sec \theta-1)}{(1+\sin \theta)} \\
& =\frac{\cot ^2 \theta(\sec \theta-1)(1-\sin \theta)(\sec \theta+1)}{(1+\sin \theta)(1-\sin \theta)(\sec \theta+1)} \\
& =\frac{\cot ^2 \theta(\sec \theta-1)(\sec \theta+1)(1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)(\sec \theta+1)} \\
& =\frac{\cot ^2 \theta\left(\sec ^2 \theta-1\right)(1-\sin \theta)}{\left(1-\sin ^2 \theta\right)(1+\sec \theta)} \\
& =\frac{\cot ^2 \theta\left(\tan ^2 \theta\right)(1-\sin \theta)}{\left(\cos ^2 \theta\right)(1+\sec \theta)} \quad\left(\because \tan ^2 \theta=\sec ^2 \theta-1,1-\sin ^2 \theta=\cos ^2 \theta\right) \\
& =\frac{\left(\cot ^\theta \tan ^2\right)}{\left(\cos ^2 \theta\right)(1+\sin \theta)} \\
& =\frac{1(1-\sin \theta)}{\left(\cos ^2 \theta\right)(1+\sec \theta)} \quad(\because \cot \theta \tan \theta=1) \\
& =\sec ^2 \theta\left(\frac{1-\sin \theta}{1+\sec \theta}\right)
\end{aligned}
$
View full question & answer→Question 274 Marks
Prove the following identity :
$
\left[\frac{1}{\left(\sec ^2 \theta-\cos ^2 \theta\right)}+\frac{1}{\left(\operatorname{cosec} 2-\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)=\frac{1-\sin ^2 \theta \cos ^2 \theta}{2+\sin ^2 \theta \cos ^2 \theta}
$
Answer$
\begin{aligned}
& \text { LHS }=\left[\frac{1}{\left(\sec ^2 \theta-\cos ^2 \theta\right)}+\frac{1}{\left(\operatorname{cosec} 2-\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right) \\
& =\left[\frac{1}{\left(\frac{1}{\cos ^2 \theta}-\cos ^2 \theta\right)}+\frac{1}{\left(\frac{1}{\sin ^2 \theta}-\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)
\end{aligned}
$
$
\begin{aligned}
& =\left[\frac{1}{\left(\frac{1-\cos ^4 \theta}{\cos ^2 \theta}\right)}+\frac{1}{\left(\frac{1-\sin ^4 \theta}{\sin ^2 \theta}\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right) \\
& \left.=\left[\frac{\cos ^2 \theta}{1-\cos ^4 \theta}+\frac{\sin ^2 \theta}{1-\sin ^4 \theta}\right)\right]\left(\sin ^2 \theta \cos ^2 \theta\right)
\end{aligned}
$
$
\begin{aligned}
& \left.=\left[\frac{\cos ^2 \theta-\cos ^2 \theta \sin ^2 \theta+\sin ^2 \theta-\sin ^2 \theta \cos ^4 \theta}{\left(1-\cos ^4 \theta\right)\left(1-\sin ^4 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)\right] \\
& =\left[\frac{\cos ^2 \theta+\sin ^2 \theta-\cos ^2 \theta \sin ^2 \theta\left(\cos ^2 \theta+\sin ^2 \theta\right)}{\left(1-\cos ^2 \theta\right)\left(1+\cos ^2 \theta\right)\left(1-\sin ^2 \theta\right)\left(1+\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right) \\
& =\left[\frac{1-\cos ^2 \theta \sin ^2 \theta}{\sin ^2 \theta\left(1+\cos ^2 \theta\right) \cos ^2 \theta\left(1+\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)
\end{aligned}
$
$
\begin{aligned}
& \left(\because \cos ^2 \theta+\sin ^2 \theta=1,\left(1-\cos ^2 \theta\right)=\sin ^2 \theta,\left(1-\sin ^2 \theta\right)=\cos ^2 \theta\right) \\
& =\frac{1-\cos ^2 \theta \sin ^2 \theta}{\left(1+\cos ^2 \theta\right)\left(1+\sin ^2 \theta\right)}=\frac{1-\cos ^2 \theta \sin ^2 \theta}{1+\sin ^2 \theta+\cos ^2 \theta+\sin ^2 \theta \cos ^2 \theta} \\
& =\frac{1-\cos ^2 \theta \sin ^2 \theta}{1+1+\sin ^2 \theta \cos ^2 \theta}=\frac{1-\sin ^2 \theta \cos ^2 \theta}{2+\sin ^2 \theta \cos ^2 \theta}
\end{aligned}
$
View full question & answer→Question 284 Marks
Prove the following identity:
$
\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta}=2
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta} \\
& =\frac{\left(\cos ^3 \theta+\sin ^3 \theta\right)(\cos \theta-\sin \theta)+\left(\cos ^3 \theta-\sin ^3 \theta\right)(\cos \theta+\sin \theta)}{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)} \\
& =\frac{\cos ^4 \theta-\cos ^3 \theta \sin \theta+\sin ^3 \theta \cos \theta-\sin ^4 \theta+\cos ^4 \theta+\cos ^3 \theta \sin \theta-\sin ^3 \theta \cos \theta-\sin ^4 \theta}{\cos ^2 \theta-\sin ^2 \theta} \\
& =\frac{2 \cos ^4 \theta-2 \sin ^4 \theta}{\cos ^2 \theta-\sin ^2 \theta}=\frac{2\left(\cos ^4 \theta-\sin ^4 \theta\right)}{\cos ^2 \theta-\sin ^2 \theta} \\
& =\frac{2\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^2 \theta-\sin ^2 \theta\right)}{\left(\cos ^2 \theta-\sin ^2 \theta\right)}=2\left(\cos ^2 \theta+\sin ^2 \theta\right) \\
& =2
\end{aligned}
$
OR
$
\begin{aligned}
& \text { LHS }=\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta} \\
& =\frac{(\cos \theta+\sin \theta)\left(\cos ^2 \theta+\sin ^2 \theta-\cos \theta \sin \theta\right)}{\cos \theta+\sin \theta}+\frac{(\cos \theta-\sin \theta)\left(\cos ^2 \theta+\sin ^2 \theta+\cos \theta \sin \theta\right)}{(\cos \theta-\sin \theta)} \\
& \left(\because a^3 \pm b^3=(a \pm b)\left(a^2+b^2 \pm a b\right)\right) \\
& =\left(\cos ^2 \theta+\sin ^2 \theta-\cos \theta \sin \theta\right)+\left(\cos ^2 \theta+\sin ^2 \theta+\cos \theta \sin \theta\right) \\
& =1-\cos \theta \sin \theta+1+\cos \theta \sin \theta \quad\left(\because \cos ^2 \theta+\sin ^2 \theta=1\right) \\
& =2
\end{aligned}
$
View full question & answer→Question 294 Marks
Prove the following identity:
$
\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin A}=2
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin A} \\
& =\frac{\left(\cos ^3 A+\sin ^3 A\right)(\cos A-\sin A)+\left(\cos ^3 A-\sin ^3 A\right)(\cos A+\sin A)}{\cos ^2 A-\sin ^2 A}
\end{aligned}
$
$
\begin{aligned}
& =\frac{\cos ^4 A-\cos ^3 A \sin A+\sin ^3 A \cos A-\sin ^4 A+\cos ^4 A+\cos ^3 A \sin A-\sin ^3 A \cos A=\sin ^4 A}{\cos ^2 A-\sin ^2 A} \\
& =\frac{2\left(\cos ^4 A-\sin ^4 A\right)}{\cos ^2 A-\sin ^2 A}=\frac{2\left(\cos ^2 A+\sin ^2 A\right)\left(\cos ^2 A-\sin ^2 A\right)}{\left(\cos ^2 A-\sin ^2 A\right)}=2\left(\cos ^2 A+\sin ^2 A\right) \\
& =2\left(\because \cos ^2 A+\sin ^2 A=1\right)
\end{aligned}
$
OR
$
\begin{aligned}
& =\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin A} \\
& = \frac{(\cos A+\sin A)\left(\cos ^2 A+\sin ^2 A-\cos A \sin A\right)}{(\cos A+\sin A)}+\frac{(\cos A-\sin A)\left(\cos ^2 A+\sin ^2 A+\cos A \sin A\right)}{(\cos A-\sin A)} \\
& \left(\because \mathrm{a}^3 \pm \mathrm{b}^3=(\mathrm{a} \pm \mathrm{b})\left(\mathrm{a}^2+\mathrm{b}^2 \pm \mathrm{ab}\right)\right) \\
& =\left(\cos ^2 A+\sin ^2 A-\cos A \sin A\right)+\left(\cos ^2 A+\sin ^2 A+\cos A \sin A\right) \\
& =1-\cos A \sin A+1+\cos A \sin A\left(\because \cos ^2 A+\sin ^2 A=1\right) \\
& =2
\end{aligned}
$
View full question & answer→