Question 13 Marks
$
\text { Prove that } \sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right)=\frac{\tan A}{1+\tan ^2 A}
$
Answer$
\begin{aligned}
& \text { LHS }=\sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right) \\
& \Rightarrow \cos A \cdot \sin A \\
& \text { RHS }=\frac{\tan A}{1+\tan ^2 A}=\frac{\tan A}{\sec ^2 A}=\frac{\frac{\sin A}{\cos A}}{\frac{1}{\cos ^2 A}} \\
& \Rightarrow \text { RHS }=\frac{\sin A}{\cos A} \cdot \cos ^2 A=\cos A \cdot \sin A
\end{aligned}
$
Thus , $\mathrm{LHS}=\mathrm{RHS}$
$
\Rightarrow \sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right)=\frac{\tan A}{1+\tan ^2 A}
$
View full question & answer→Question 23 Marks
For $\triangle A B C$, prove that:
$
\sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}
$
Answer$
\sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}
$
We know that for a triangle $\triangle A B C$
$
\begin{aligned}
& <A+<B+<C=180^{\circ} \\
& \frac{<B+<A}{2}=90^{\circ}-\frac{<C}{2} \\
& \sin \left(\frac{A+B}{2}\right)=\sin \left(90^{\circ}-\frac{C}{2}\right) \\
& =\cos \left(\frac{C}{2}\right)
\end{aligned}
$
View full question & answer→Question 33 Marks
For $\triangle A B C$, prove that :
$
\tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2}
$
Answer$
\tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2}
$
We know that for a triangle $\triangle A B C$
$
\begin{aligned}
& <A+<B+<C=180^{\circ} \\
& <B+<C=180^{\circ}-<A \\
& \frac{<B+<C}{2}=90^{\circ}-\frac{<A}{2} \\
& \tan \left(\frac{B+C}{2}\right)=\tan \left(90^{\circ}-\frac{A}{2}\right) \\
& =\cot \left(\frac{A}{2}\right)
\end{aligned}
$
View full question & answer→Question 43 Marks
Find the value of $\theta\left(0^{\circ}<\theta<90^{\circ}\right)$ if :
$
\tan 35^{\circ} \cot \left(90^{\circ}-\theta\right)=1
$
Answer$
\begin{aligned}
& \tan 35^{\circ} \cot \left(90^{\circ}-\theta\right)=1 \\
& \text { Given } \tan 35^{\circ} \cot \left(90^{\circ}-\theta\right)=1 \\
& \Rightarrow \tan 35^{\circ} \tan \theta=1 \\
& \Rightarrow \tan 35^{\circ}=\cot \theta \\
& \Rightarrow \tan 35^{\circ}=\tan \left(90^{\circ}-\theta\right) \\
& \Rightarrow 90^{\circ}-\theta=35^{\circ} \\
& \Rightarrow \theta=55^{\circ}
\end{aligned}
$
View full question & answer→Question 53 Marks
Without using trigonometric table, evaluate:
$
\left(\frac{\sin 47^{\circ}}{\cos 43^{\circ}}\right)^2-4 \cos ^2 45^{\circ}+\left(\frac{\cos 43^{\circ}}{\sin 47^{\circ}}\right)^2
$
Answer$
\begin{aligned}
& \left(\frac{\sin 47^{\circ}}{\cos 43^{\circ}}\right)^2-4 \cos ^2 45^{\circ}+\left(\frac{\cos 43^{\circ}}{\sin 47^{\circ}}\right)^2 \\
& \Rightarrow\left(\frac{\sin 47^{\circ}}{\cos 43^{\circ}}\right)^2+\left(\frac{\cos 43^{\circ}}{\sin 47^{\circ}}\right)^2-4\left(\frac{1}{\sqrt{2}}\right)^2 \\
& \Rightarrow\left(\frac{\sin \left(90^{\circ}-43^{\circ}\right)}{\cos 43^{\circ}}\right)^2+\left(\frac{\cos \left(90^{\circ}-47^{\circ}\right)}{\sin 47^{\circ}}\right)^2-4\left(\frac{1}{2}\right) \\
& \Rightarrow\left(\frac{\cos 43^{\circ}}{\cos 43^{\circ}}\right)^2+\left(\frac{\sin 47^{\circ}}{\sin 47^{\circ}}\right)^2-2 \\
& \Rightarrow 1+1-2=0
\end{aligned}
$
View full question & answer→Question 63 Marks
$
\text { prove that } \frac{1}{1+\cos \left(90^{\circ}-A\right)}+\frac{1}{1-\cos \left(90^{\circ}-A\right)}=2 \operatorname{cosec} c^2\left(90^{\circ}-A\right)
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{1}{1+\cos \left(90^{\circ}-A\right)}+\frac{1}{1-\cos \left(90^{\circ}-A\right)} \\
& =\frac{1}{1+\sin A}+\frac{1}{1-\sin A} \\
& =\frac{1-\sin A+1+\sin A}{(1+\sin A)(1-\sin A)} \\
& =\frac{2}{1-\sin ^2 A} \\
& =\frac{2}{\cos ^2 A} \\
& =2 \sec ^2 A=2 \operatorname{cosec}\left(90^{\circ}-A\right)
\end{aligned}
$
View full question & answer→Question 73 Marks
If $sinA + cosA = m$ and $secA + cosecA = n$, prove that $n(m^2 - 1) = 2m$
AnswerGiven : $\sin \theta+\cos \theta=m$ and $\sec \theta+\operatorname{cosec} \theta=n$
$
\begin{aligned}
& \text { Consider L.H.S. }=n\left(m^2-1\right)=(\sec \theta+\operatorname{cosec} \theta)\left[(\sin \theta+\cos \theta)^2-1\right] \\
& =\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)\left[\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta-1\right] \\
& =\left(\frac{\cos \theta+\sin \theta}{\sin \theta \cos \theta}\right)(1+2 \sin \theta \cos \theta-1) \\
& =\frac{(\cos \theta+\sin \theta)}{\sin \theta \cos \theta}(2 \sin \theta \cos \theta) \\
& =2(\sin \theta+\cos \theta) \\
& =2 m=\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 83 Marks
$
\text { If } \mathrm{x}=\mathrm{r} \sin \mathrm{A} \cos \mathrm{B}, \mathrm{y}=\mathrm{r} \sin \mathrm{A} \sin \mathrm{B} \text { and } \mathrm{z}=\mathrm{r} \cos \mathrm{A} \text {, prove that } x^2+y^2+z^2=r^2
$
Answer$
\begin{aligned}
& \text { LHS }=(r \sin A \cos B)^2+(r \sin A \sin B)^2+(r \cos A)^2 \\
& \Rightarrow r^2 \sin ^2 A \cos ^2 B+r^2 \sin ^2 A \sin ^2 B+r^2 \cos ^2 A \\
& \Rightarrow r^2 \sin ^2 A\left(\cos ^2 B+\sin ^2 B\right)+r^2 \cos ^2 A \\
& \Rightarrow r^2\left(\sin ^2 A+\cos ^2 A\right)=r^2=\text { RHS }
\end{aligned}
$
View full question & answer→Question 93 Marks
$
\text { If } a \sin ^2 \theta+b \cos ^2 \theta=c \text { and } p \sin ^2 \theta+q \cos ^2 \theta=r \text {, prove that }(b-c)(r-p)=(c-a)(q-r)
$
Answer$
\begin{aligned}
& \text { LHS }=(b-c)(r-p)=\left(b-a \sin ^2 \theta-b \cos ^2 \theta\right)\left(p \sin ^2 \theta+q \cos ^2 \theta-p\right) \\
& =\left[b\left(1-\cos ^2 \theta\right)-a \sin ^2 \theta\right]\left[p\left(\sin ^2 \theta-1\right)+q \cos ^2 \theta\right] \\
& \Rightarrow \text { LHS }=\left[(b-a) \sin ^2 \theta\right]\left[(q-p) \cos ^2 \theta\right]=(b-a)(q-p) \sin ^2 \theta \cos ^2 \theta \\
& \text { RHS }=(c-a)(q-r)=\left(a \sin ^2 \theta+b \cos ^2 \theta-a\right)\left(q-p \sin ^2 \theta-q \cos ^2 \theta\right) \\
& =\left[(b-a) \cos ^2 \theta\right]\left[(q-p) \sin ^2 \theta\right]=(b-a)(q-p) \sin ^2 \theta \cdot \cos ^2 \theta
\end{aligned}
$
Thus, $(b-c)(r-p)=(c-a)(q-r)$
View full question & answer→Question 103 Marks
If $m = a\ secA + b\ tanA$ and $n = a\ tanA + b\ secA$ , prove that $m^2 - n^2 = a^2 - b^2$
Answer$
\begin{aligned}
& \text { Given, } \mathrm{m}=\mathrm{a} \sec \mathrm{A}+\mathrm{b} \tan \mathrm{A} \text { and } \mathrm{n}=\mathrm{a} \tan \mathrm{A}+\mathrm{b} \sec \mathrm{A} \\
& m^2-n^2=(a \sec A+b \tan A)^2-(a \tan A+b \sec A)^2 \\
& \Rightarrow a^2 \sec ^2 A+b^2 \tan ^2 A+2 a b \sec A \tan A-\left(a^2 \tan ^2 A+b^2 \sec ^2 A+2 a b \sec A \tan A\right) \\
& \Rightarrow \sec ^2 A\left(a^2-b^2\right)+\tan ^2 A\left(b^2-a^2\right)=\left(a^2-b^2\right)\left[\sec ^2 A-\tan ^2 A\right] \\
& \Rightarrow\left(a^2-b^2\right)\left[\text { Since } \sec ^2 A-\tan ^2 A=1\right]
\end{aligned}
$
Hence, $m^2-n^2=a^2-b^2$
View full question & answer→Question 113 Marks
Prove the following identity:
$
\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\frac{\sec \theta+1}{\sec \theta-1}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta} \\
& =\frac{\frac{\sin \theta}{\cos \theta}+\sin \theta}{\frac{\sin \theta}{\cos \theta}-\sin \theta}=\frac{\sin \theta+\sin \theta \cos \theta}{\sin \theta+\sin \theta \cos \theta} \\
& =\frac{\sin \theta(1+\cos \theta)}{\sin (1+\cos \theta)}=\frac{1+\cos \theta}{1-\cos \theta} \\
& =\frac{1+\frac{1}{\sec \theta}}{1-\frac{1}{\sec \theta}}=\frac{\frac{\sec \theta+1}{\sec \theta}}{\frac{\sec \theta-1}{\sec \theta}} \\
& =\frac{\sec \theta+1}{\sec \theta-1}
\end{aligned}
$
View full question & answer→Question 123 Marks
Prove the following identity:
$
\frac{1+\sin \theta}{\operatorname{cosec} \theta-\cot \theta}-\frac{1-\sin \theta}{\operatorname{cosec} \theta+\cot \theta}=2(1+\cot \theta)
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{1+\sin \theta}{\operatorname{cosec} \theta-\cot \theta}-\frac{1-\sin \theta}{\operatorname{cosec} \theta+\cot \theta} \\
& =\frac{(1+\sin \theta)(\operatorname{cosec} \theta+\cot \theta)-(1-\sin \theta)(\operatorname{cosec} \theta-\cot \theta)}{\operatorname{cosec} 2 \theta-\cot ^2 \theta} \\
& =\frac{\operatorname{cosec} \theta+\cot \theta+1+\cos \theta-\operatorname{cosec} \theta+\cot \theta+1-\cos \theta}{1+\cot ^2 \theta-\cot ^2 \theta} \quad\left(\because \operatorname{cosec}{ }^2 \theta=1+\cot ^2 \theta\right) \\
& =2+2 \cot \theta=2(1+\cot \theta)
\end{aligned}
$
View full question & answer→Question 133 Marks
Prove the following identity:
$
\left(1+\tan ^2 \theta\right) \sin \theta \cos \theta=\tan \theta
$
Answer$
\begin{aligned}
& \text { LHS }=\left(1+\tan ^2 \theta\right) \sin \theta \cos \theta \\
& =\left(1+\frac{\sin ^2 \theta}{\cos ^2 \theta}\right) \sin \theta \cos \theta \\
& =\left(\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}\right) \sin \theta \cos \theta \\
& =\frac{1}{\cos ^2 \theta} \times \sin \theta \cos \theta \quad\left(\because \cos ^2 \theta+\sin 2 \theta=1\right) \\
& =\frac{\sin \theta}{\cos \theta}=\tan \theta
\end{aligned}
$
View full question & answer→Question 143 Marks
Prove the following identity :
$
\frac{\operatorname{cosec} \theta}{\tan \theta+\cot \theta}=\cos \theta
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\cos e c \theta}{\tan \theta+\cot \theta} \\
& =\frac{\frac{1}{\sin \theta}}{\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}} \\
& =\frac{\frac{1}{\sin \theta}}{\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}}=\frac{\frac{1}{\sin \theta}}{\frac{1}{\cos \theta \sin \theta}} \\
& =\frac{1}{\sin \theta} \times \frac{\cos \theta \sin \theta}{1}=\cos \theta
\end{aligned}
$
View full question & answer→Question 153 Marks
Prove the following identity:
$
\frac{\sec A-1}{\sec A+1}=\frac{\sin ^2 A}{(1+\cos A)^2}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\sec A-1}{\sec A+1} \\
& =\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1}=\frac{1-\cos A}{1+\cos A} \\
& =\frac{1-\cos A}{1+\cos A} \times \frac{1+\cos A}{1+\cos A} \\
& =\frac{1-\cos ^2 A}{(1+\cos A)^2} \\
& =\frac{\sin ^2 A}{(1+\cos A)^2} \quad\left(\because 1-\cos ^2 A=\sin ^2 A\right)
\end{aligned}
$
View full question & answer→Question 163 Marks
Prove the following identity:
$
\left(1+\tan ^2 A\right)+\left(1+\frac{1}{\tan ^2 A}\right)=\frac{1}{\sin ^2 A-\sin ^4 A}
$
Answer$
\begin{aligned}
& \text { LHS }=\left(1+\tan ^2 A\right)+\left(1+\frac{1}{\tan ^2 A}\right) \\
& =\left(1+\frac{\sin ^2 A}{\cos ^2 A}\right)+\left(1+\frac{1}{\frac{\sin ^2 A}{\cos ^2 A}}\right) \\
& =\left(\frac{\cos ^2 A+\sin ^2 A}{\cos ^2 A}\right)+\left(\frac{\cos ^2 A+\sin ^2 A}{\sin ^2 A}\right) \\
& =\frac{1}{1-\sin ^2 A}+\frac{1}{\sin ^2 A}\left(\because \cos ^2 A+\sin ^2 A=1\right) \\
& =\frac{\sin ^2 A+1-\sin ^2 A}{\sin ^2 A\left(1-\sin ^2 A\right)}=\frac{1}{\sin ^2 A-\sin ^4 A}
\end{aligned}
$
View full question & answer→Question 173 Marks
Prove the following identity :
$
(\sec \theta-\tan \theta)^2=\frac{1-\sin \theta}{1+\sin \theta}
$
Answer$
\begin{aligned}
& (\sec \theta-\tan \theta)^2 \\
& =\left(\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\right)^2 \\
& =\left(\frac{1-\sin \theta}{\cos \theta}\right)^2=\frac{(1-\sin \theta)^2}{\cos ^2 \theta} \\
& =\frac{(1-\sin \theta)^2}{1-\sin ^2 \theta}=\frac{(1-\sin \theta)^2}{(1-\sin \theta)(1+\sin \theta)}\left(\because 1-\sin ^2 \theta=\cos ^2 \theta\right. \\
& =\frac{1-\sin \theta}{1+\sin \theta}
\end{aligned}
$
View full question & answer→Question 183 Marks
Prove the following identity:
$
\sqrt{\frac{1+\cos A}{1-\cos A}}=\operatorname{cosec} A+\cot A
$
Answer$
\begin{aligned}
& =\sqrt{\frac{1+\cos A}{1-\cos A} \cdot \frac{1+\cos A}{1+\cos A}} \\
& =\sqrt{\frac{(1+\cos A)^2}{1-\cos ^2 A}}=\sqrt{\frac{(1+\cos A)^2}{\sin ^2 A}} \\
& =\sqrt{\frac{1+\cos ^2 A}{\sin A}}=\sqrt{\frac{1}{\sin A}+\frac{\cos ^2 A}{\sin A}} \\
& =\sqrt{\left(\operatorname{cosec} A+\cot ^2 A\right)} \\
& =\operatorname{cosec} A+\cot A
\end{aligned}
$
View full question & answer→Question 193 Marks
Prove the following identity:
$
\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}
$
Answer$
\begin{aligned}
& =\text { LHS }=\sqrt{\frac{1-\cos A}{1-\cos A}} \\
& =\sqrt{\frac{1-\cos A}{1+\cos A} \cdot \frac{1+\cos A}{1+\cos A}} \\
& =\sqrt{\frac{1-\cos ^2 A}{(1+\cos A)^2}} \\
& =\sqrt{\frac{\sin ^2 A}{(1+\cos A)^2}} \\
& =\frac{\sin A}{1+\cos A}
\end{aligned}
$
View full question & answer→Question 203 Marks
Prove the following identity:
$
\frac{1+\cos A}{1-\cos A}=\frac{\tan ^2 A}{(\sec A-1)^2}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{1+\cos A}{1-\cos A} \\
& =\frac{1+\frac{1}{\sec A}}{1-\frac{1}{\sec A}}=\frac{\sec A+1}{\sec A-1} \\
& =\frac{\sec A+1}{\sec A-1} \cdot \frac{\sec A-1}{\sec A-1} \\
& =\frac{\sec ^2 A-1}{(\sec A-1)^2}=\frac{\tan ^2 A}{(\sec A-1)^2}\left(Q \sec ^2 A-1=\tan ^2 A\right)
\end{aligned}
$
View full question & answer→Question 213 Marks
Prove the following identity:
$
\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}=2 \sec ^2 A
$
Answer$
\begin{aligned}
& \frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}=2 \sec ^2 A \\
& \text { LHS }=\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1} \\
& =\frac{\operatorname{cosec}{ }^2 A+\operatorname{cosec} A+\operatorname{cosec}{ }^2 A-\operatorname{cosec} A}{\operatorname{cosec} A-1} \\
& =\frac{2 \operatorname{cosec}{ }^2 A}{\cot ^2 A}\left(Q \operatorname{cosec}{ }^2 A-1=\cot ^2 A\right) \\
& =\frac{\frac{2}{\sin ^2 A}}{\frac{\cos ^2 A}{\sin ^2 A}}=\frac{2}{\cos ^2 A}=2 \sec ^2 A
\end{aligned}
$
View full question & answer→Question 223 Marks
Prove the following identity:
$
\operatorname{cosec} A+\cot A=\frac{1}{\operatorname{cosec} A-\cot A}
$
Answer$
\begin{aligned}
& \text { LHS }=\operatorname{cosec} A+\cot A \\
& =\frac{\operatorname{cosec} A+\cot A}{1} \cdot \frac{\operatorname{cosec} A-\cot A}{\operatorname{cosec} A-\cot A} \\
& =\frac{\operatorname{cosec}{ }^2 A-\cot ^2 A}{\operatorname{cosec} A-\cot A}=\frac{1+\cot ^2 A-\cot ^2 A}{\operatorname{cosec} A-\cot A} \\
& =\frac{1}{\operatorname{cosec} A-\cot A}
\end{aligned}
$
View full question & answer→Question 233 Marks
Prove the following identity:
$
\frac{\left(1+\tan ^2 A\right) \cot A}{\operatorname{cosec} 2}=\tan A
$
Answer$
\begin{aligned}
& \frac{\left(1+\tan ^2 A\right) \cot A}{\operatorname{cosec} A} \\
& =\frac{\sec ^2 A \cot A}{\operatorname{cosec}^2 A} \ldots \ldots .\left(\therefore \sec ^2 A=1+\tan ^2 A\right) \\
& =\frac{\frac{1}{\cos ^2 A} \cdot \frac{\cos A}{\sin A}}{\frac{1}{\sin ^2 A}}=\frac{1}{\frac{\cos A \sin A}{\frac{1}{\sin ^2 A}}} \\
& =\frac{\sin A}{\cos A}=\tan A
\end{aligned}
$
View full question & answer→Question 243 Marks
Prove the following identity:
$
\tan A-\cot A=\frac{1-2 \cos ^2 A}{\sin A \cos A}
$
Answer$
\begin{aligned}
& \tan A-\cot A=\frac{\sin A}{\cos A}-\frac{\cos A}{\sin A} \\
& =\frac{\sin ^2 A-\cos ^2 A}{\sin A \cos A} \\
& =\frac{1-\cos ^2 A-\cos ^2 A}{\sin A \cos A}\left(Q \sin ^2 A=1-\cos ^2 A\right) \\
& =\frac{1-2 \cos ^2 A}{\sin A \cos A}
\end{aligned}
$
View full question & answer→Question 253 Marks
Prove the following identity:
$
\frac{1}{\tan A+\cot A}=\sin A \cos A
$
Answer$
\begin{aligned}
& \frac{1}{\tan A+\cot A}=\sin A \cos A \\
& \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 A \cos A}} \\
& =\frac{1}{\frac{1}{\sin A \cos A}}\left(Q \sin ^2 A+\cos ^2 A=1\right) \\
& =\sin A \cos A
\end{aligned}
$
View full question & answer→Question 263 Marks
Prove the following identity:
$
\frac{\cot A+\tan B}{\cot B+\tan A}=\cot A \tan B
$
Answer$
\begin{aligned}
& \frac{\cot A+\tan B}{\cot B+\tan A}=\cot A \tan B \\
& \frac{\cot A+\tan B}{\cot B+\tan A} \\
& =\frac{\frac{1}{\tan A}+\tan B}{\frac{1}{\tan B}+\tan A} \\
& =\frac{\frac{1+\tan A \tan B}{\tan A}}{\frac{1+\tan A \tan B}{\tan B}}=\frac{1+\tan A \tan B}{\tan A} \cdot \frac{\tan B}{1+\tan A \tan B} \\
& =\frac{\tan B}{\tan A}=\frac{1}{\tan A} \cdot \tan B=\cot A \tan B
\end{aligned}
$
View full question & answer→Question 273 Marks
Prove the following identity:
$
\frac{1+\cos A}{1-\cos A}=(\operatorname{cosec} A+\cot A)^2
$
Answer$
\begin{aligned}
& \frac{1+\cos A}{1-\cos A}=(\operatorname{cosec} A+\cot A)^2 \\
& =\frac{1+\cos A}{1-\cos A} \cdot \frac{1+\cos A}{1+\cos A} \\
& =\frac{(1+\cos A)^2}{1-\cos ^2 A}=\frac{(1+\cos A)^2}{\sin ^2 A} \\
& =\left[\frac{1+\cos A}{\sin A}\right]^2=\left[\frac{1}{\sin A}+\frac{\cos A}{\sin A}\right]^2 \\
& =(\operatorname{cosec} A+\cot A)^2
\end{aligned}
$
View full question & answer→Question 283 Marks
Prove the following identity:
$
\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=2 \operatorname{cosec} A
$
Answer$
\begin{aligned}
& \frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=2 \operatorname{cosec} A \\
& \frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A} \\
& =\frac{(1+\cos A)^2+\sin ^2 A}{\sin A(1+\cos A)} \\
& =\frac{1+2 \cos A+\cos ^2 A+\sin ^2 A}{\sin A(1+\cos A)} \\
& =\frac{2+2 \cos A}{\sin A(1+\cos A)} \\
& =\frac{2(1+\cos A)}{\sin A(1+\cos A)} \quad\left[\sin ^2 A+\cos ^2 A=1\right] \\
& =2 \operatorname{cosec} A
\end{aligned}
$
View full question & answer→Question 293 Marks
Prove the following Identities:
$
\frac{\operatorname{cosec} A}{\cot A+\tan A}=\cos A
$
Answer$
\begin{aligned}
& \frac{\operatorname{cosec} A}{\cot A+\tan A}=\cos A \\
& =\mathrm{LHS} \\
& =\frac{\operatorname{cosec} A}{\cot A+\tan A} \\
& =\frac{\operatorname{cosec} A}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}} \\
& =\frac{\frac{\operatorname{cosec} A}{\cos A+\sin ^2 A}}{\sin A \cdot \cos A} \\
& =\frac{\frac{1}{\sin A}}{\frac{1}{\sin A \cdot \cos A}} \\
& =\frac{\sin A \cdot \cos A}{\sin A} \\
& =\cos A \\
& =\text { RHS }
\end{aligned}
$
View full question & answer→Question 303 Marks
Prove the following identity:
$
\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A} \\
& =\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}} \\
& =\frac{\cos ^2 A}{\cos A-\sin A}+\frac{\sin ^2 A}{\sin A-\cos A}=\frac{\cos ^2 A-\sin ^2 A}{(\cos A-\sin A)} \\
& \frac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A} \\
& =\sin A+\cos A=\text { RHS }
\end{aligned}
$
View full question & answer→Question 313 Marks
Prove the following identity:
$
\sec ^2 A \cdot \operatorname{cosec}{ }^2 A=\tan ^2 A+\cot ^2 A+2
$
Answer$
\begin{aligned}
& \text { LHS }=\sec ^2 A \cdot \operatorname{cosec} c^2 A=\frac{1}{\cos ^2 A \cdot \sin ^2 A} \\
& \text { RHS }=\tan ^2 A+\cot ^2 A+2=\tan ^2 A+\cot ^2 A+2 \tan ^2 A \cdot \cot ^2 A \\
& =(\tan A+\cot A)^2=\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)^2 \\
& =\left(\frac{\sin ^2 A+\cos ^2 A}{\sin A \cdot \cos A}\right)^2=\frac{1}{\cos ^2 A \cdot \sin ^2 A} \\
& =\text { Hence, LHS = RHS }
\end{aligned}
$
View full question & answer→Question 323 Marks
Prove the following identity:
$
(\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1
$
Answer$
\begin{aligned}
& \text { LHS }=(\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A) \\
& =\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)\left(\tan A+\frac{1}{\tan A}\right) \\
& =\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right)\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right) \\
& =\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)\left(\frac{\sin ^2 A+\cos ^2 A}{\sin A \cdot \cos A}\right) \\
& =1
\end{aligned}
$
View full question & answer→Question 333 Marks
Prove the following identity:
$
\operatorname{cosec}{ }^4 A-\operatorname{cosec} c^2 A=\cot ^4 A+\cot ^2 A
$
Answer$
\begin{aligned}
& \text { LHS }=\operatorname{cosec} c^4 A-\operatorname{cosec} c^2 A \\
& =\operatorname{cosec} c^2 A\left(\operatorname{cosec} c^2 A-1\right) \\
& \text { RHS }=\cot ^4 A+\cot ^2 A \\
& =\cot ^2 A\left(\cot ^2 A+1\right) \\
& =\left(\operatorname{cosec} c^2 A-1\right) \operatorname{cosec} c^2 A
\end{aligned}
$
Thus, $\mathrm{LHS}=\mathrm{RHS}$
View full question & answer→Question 343 Marks
Prove the following identity:
$
\frac{\cos A}{1+\sin A}=\sec A-\tan A
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\cos A}{1+\sin A} \\
& \text { RHS }=\sec A-\tan A \\
& =\frac{1}{\cos A}-\frac{\sin A}{\cos A}=\frac{1-\sin A}{\cos A} \\
& =\frac{1-\sin A}{\cos A}\left(\frac{1+\sin A}{1+\sin A}\right)=\left(\frac{1-\sin ^2 A}{\cos A(1+\sin A)}\right) \\
& =\frac{\cos ^2 A}{\cos A(1+\sin A)}=\frac{\cos A}{(1+\sin A)}=\mathrm{LHS}
\end{aligned}
$
View full question & answer→Question 353 Marks
Prove the following identity :
sinθcotθ + sinθcosecθ = 1 + cosθ
Answer$
\begin{aligned}
& \sin \theta \cot \theta+\sin \theta \operatorname{cosec} \theta=1+\cos \theta \\
& \text { LHS }=\sin \theta \frac{\cos \theta}{\sin \theta}+\sin \theta \frac{1}{\sin \theta} \\
& =\cos \theta+1=\text { RHS }
\end{aligned}
$
View full question & answer→Question 363 Marks
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Answer$
\begin{aligned}
& (1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta) \\
& =\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{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\
& =\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2
\end{aligned}
$
View full question & answer→