Question 12 Marks
$
\text { Prove that } \frac{\sin A}{\sin \left(90^{\circ}-A\right)}+\frac{\cos A}{\cos \left(90^{\circ}-A\right)}=\sec \left(90^{\circ}-A\right) \cos e c\left(90^{\circ}-A\right)
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{\sin A}{\sin \left(90^{\circ}-A\right)}+\frac{\cos A}{\cos \left(90^{\circ}-A\right)} \\
& \Rightarrow \frac{\cos \left(90^{\circ}-A\right)}{\sin \left(90^{\circ}-A\right)}+\frac{\sin \left(90^{\circ}-A\right)}{\cos \left(90^{\circ}-A\right)} \\
& \Rightarrow \frac{\cos ^2\left(90^{\circ}-A\right)+\sin ^2\left(90^{\circ}-A\right)}{\sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right)}=\frac{1}{\sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right)} \\
& \Rightarrow \sec \left(90^{\circ}-A\right) \operatorname{cosec}\left(90^{\circ}-A\right)
\end{aligned}
$
View full question & answer→Question 22 Marks
Without using trigonometric identity, show that:
$
\sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \cos e c 70^{\circ}=0
$
Answer$
\begin{aligned}
& \sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \operatorname{cosec} 70^{\circ}=0 \\
& \text { Consider } \sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \operatorname{cosec} 70^{\circ} \\
& \Rightarrow \sec \left(90^{\circ}-20^{\circ}\right) \sin 20^{\circ}-\cos 20^{\circ} \cdot \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right) \\
& \Rightarrow \operatorname{cosec} 20^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \sec 20^{\circ} \\
& \Rightarrow \frac{1}{\sin 20^{\circ}} \cdot \sin 20^{\circ}-\cos 20^{\circ} \cdot \frac{1}{\cos 20^{\circ}} \\
& \Rightarrow 1-1=0
\end{aligned}
$
View full question & answer→Question 32 Marks
Without using trigonometric identity, show that:
$
\cos ^2 25^{\circ}+\cos ^2 65^{\circ}=1
$
Answer$
\begin{aligned}
& \text { Consider } \cos ^2 25^{\circ}+\cos ^2 65^{\circ} \\
& \Rightarrow \cos ^2\left(90^{\circ}-65^{\circ}\right)+\cos ^2 65^{\circ} \\
& \Rightarrow \sin ^2 65^{\circ}+\cos ^2 65^{\circ}=1
\end{aligned}
$
View full question & answer→Question 42 Marks
Without using trigonometric identity, show that:
$
\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)=0
$
Answer$
\begin{aligned}
& \sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)=0 \\
& \sin \left(50^{\circ}+\theta\right)=\cos \left[90^{\circ}-\left(50^{\circ}+\theta\right)\right]=\cos \left(40^{\circ}-\theta\right) \\
& \sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right) \\
& =\cos \left(40^{\circ}-\theta\right)-\cos \left(40^{\circ}-\theta\right) \\
& =0
\end{aligned}
$
View full question & answer→Question 52 Marks
Without using trigonometric identity, show that:
$
\tan 10^{\circ} \tan 20^{\circ} \tan 30^{\circ} \tan 70^{\circ} \tan 80^{\circ}=\frac{1}{\sqrt{3}}
$
Answer$
\begin{aligned}
& \tan 10^{\circ} \tan 20^{\circ} \tan 30^{\circ} \tan 70^{\circ} \tan 80^{\circ}=\frac{1}{\sqrt{3}} \\
& \text { Consider } \tan 10^{\circ} \tan 20^{\circ} \tan 30^{\circ} \tan 70^{\circ} \tan 80^{\circ} \\
& \Rightarrow \tan \left(90^{\circ}-80^{\circ}\right)-\tan \left(90^{\circ}-70^{\circ}\right) \tan 30^{\circ} \tan 70^{\circ} \tan 80^{\circ} \\
& \Rightarrow \cot 80^{\circ} \cdot \cot 70^{\circ} \cdot \tan 30^{\circ} \tan 70^{\circ} \tan 80^{\circ} \\
& \Rightarrow \tan 30^{\circ}=\frac{1}{\sqrt{3}}[\text { As } \tan \theta \cot \theta=1]
\end{aligned}
$
View full question & answer→Question 62 Marks
Without using trigonometric identity, show that :
$
\sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ}=2
$
Answer$
\begin{aligned}
& \sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ}=2 \\
& \text { consider } \sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ} \\
& \Rightarrow \sin 42^{\circ} \sec \left(90^{\circ}-42^{\circ}\right)+\cos 42^{\circ} \operatorname{cosec}\left(90^{\circ}-42^{\circ}\right) \\
& \Rightarrow \sin 42^{\circ} \cdot \operatorname{cosec} 42^{\circ}+\cos 42^{\circ} \sec 42^{\circ} \\
& \Rightarrow \sin 42^{\circ} \cdot \frac{1}{\sin 42^{\circ}}+\cos 42^{\circ} \frac{1}{\cos 42^{\circ}} \\
& \Rightarrow 1+1=2
\end{aligned}
$
View full question & answer→Question 72 Marks
Find the value of $\theta\left(0^{\circ}<\theta<90^{\circ}\right)$ if : $\cos 63^{\circ} \sec \left(90^{\circ}-\theta\right)=1$
Answer$
\begin{aligned}
& \cos 63^{\circ} \sec \left(90^{\circ}-\theta\right)=1 \\
& \cos 63^{\circ} \cos e c \theta=1 \\
& \Rightarrow \cos 63^{\circ}=\sin \theta \\
& \Rightarrow \cos 63^{\circ}=\cos \left(90^{\circ}-\theta\right) \\
& \Rightarrow 63^{\circ}=90^{\circ}-\theta \\
& \Rightarrow \theta=27^{\circ}
\end{aligned}
$
View full question & answer→Question 82 Marks
Without using trigonometric table, evaluate:
$
\frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}}
$
Answer$
\begin{aligned}
& \frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}} \\
& \Rightarrow \frac{\sin \left(90^{\circ}-18^{\circ}\right)}{\cos 18^{\circ}}-\frac{\sec \left(90^{\circ}-58^{\circ}\right)}{\cos e c 58^{\circ}} \\
& \Rightarrow \frac{\cos 18^{\circ}}{\cos 18^{\circ}}-\frac{\operatorname{cosec} 58^{\circ}}{\operatorname{cosec} 58^{\circ}}=1-1=0
\end{aligned}
$
View full question & answer→Question 92 Marks
Without using trigonometric table, evaluate:
$
\left(\frac{\sin 49^{\circ}}{\sin 41^{\circ}}\right)^2+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^2
$
Answer$
\begin{aligned}
& \left(\frac{\sin 49^{\circ}}{\sin 41^{\circ}}\right)^2+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^2 \\
& \Rightarrow\left(\frac{\sin \left(90^{\circ}-41^{\circ}\right)}{\sin 41^{\circ}}\right)^2+\left(\frac{\cos \left(90^{\circ}-49^{\circ}\right)}{\sin 49^{\circ}}\right)^2 \\
& \Rightarrow\left(\frac{\cos 41^{\circ}}{\sin 41^{\circ}}\right)^2+\left(\frac{\sin 49^{\circ}}{\sin 49^{\circ}}\right)^2 \\
& \Rightarrow 1+1=2
\end{aligned}
$
View full question & answer→Question 102 Marks
Without using trigonometric table, evaluate:
$
\cos 90^{\circ}+\sin 30^{\circ} \tan 45^{\circ} \cos ^2 45^{\circ}
$
Answer$
\begin{aligned}
& \cos 90^{\circ}+\sin 30^{\circ} \tan 45^{\circ} \cos ^2 45^{\circ} \\
& \Rightarrow \cos 90^{\circ}+\sin 30^{\circ} \cdot \frac{\sin 45^{\circ}}{\cos 45^{\circ}} \cdot \cos ^2 45^{\circ} \\
& \Rightarrow \cos 90^{\circ}+\sin 30^{\circ} \cdot \sin 45^{\circ} \cdot \cos 45^{\circ} \\
& \Rightarrow 0+\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}
\end{aligned}
$
View full question & answer→Question 112 Marks
Without using trigonometric table, evaluate:
$
\cos e c 49^{\circ} \cos 41^{\circ}+\frac{\tan 31^{\circ}}{\cot 59^{\circ}}
$
Answer$
\begin{aligned}
& \operatorname{cosec} 49^{\circ} \cos 41^{\circ}+\frac{\tan 31^{\circ}}{\cot 59^{\circ}} \\
& \Rightarrow \sec \left(90^{\circ}-41^{\circ}\right) \cos 41^{\circ}+\frac{\cot \left(90^{\circ}-59^{\circ}\right)}{\cot 56^{\circ}} \\
& \Rightarrow \sec 41^{\circ} \cos 41^{\circ}+\frac{\cot 59^{\circ}}{\cot 59^{\circ}}=1+1=2
\end{aligned}
$
View full question & answer→Question 122 Marks
$
\text { If } \sin A+\cos A=\sqrt{2} \text {, prove that } \sin A \cos A=\frac{1}{2}
$
AnswerWe Know, $(\sin A+\cos A)^2=\sin ^2 A+\cos ^2 A+2 \sin A \cdot \cos A$ Given, $(\sin A+\cos A)=\sqrt{2}$
$
\begin{aligned}
& \Rightarrow 2=1+2 \sin \mathrm{A} \cdot \cos \mathrm{A} \\
& \Rightarrow 2 \sin \mathrm{A} \cdot \cos \mathrm{A}=1 \\
& \Rightarrow \sin \mathrm{A} \cdot \cos \mathrm{A}=\frac{1}{2}
\end{aligned}
$
View full question & answer→Question 132 Marks
$
\text { If } \mathrm{x}=\mathrm{asec} \theta+\mathrm{b} \tan \theta \text { and } \mathrm{y}=\operatorname{atan} \theta+\mathrm{bsec} \theta \text {, prove that } x^2-y^2=a^2-b^2
$
Answer$
\begin{aligned}
& x^2-y^2=(a \sec \theta+b \operatorname{Tan} \theta)^2-(a \operatorname{Tan} \theta+b \operatorname{Sec} \theta)^2 \\
& \Rightarrow a^2 \sec ^2 \theta+b^2 \operatorname{Tan}^2 \theta+2 a b \operatorname{Sec} \theta \operatorname{Tan} \theta-\left(a^2 \operatorname{Tan}^2 \theta+b^2 \operatorname{Sec}^2 \theta+2 a b \operatorname{Sec} \theta \operatorname{Tan} \theta\right) \\
& \Rightarrow \sec ^2 \theta\left(a^2-b^2\right)+\operatorname{Tan}^2 \theta\left(b^2-a^2\right)=\left(a^2-b^2\right)\left[\operatorname{Sec}^2 \theta-\operatorname{Tan}^2 \theta\right] \\
& \Rightarrow\left(a^2-b^2\right) \quad\left[\text { Since } \sec ^2 \theta-\operatorname{Tan}^2 \theta=1\right]
\end{aligned}
$
Hence, $x^2-y^2=a^2-b^2$
View full question & answer→Question 142 Marks
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Answer$
\begin{aligned}
& \text { LHS }=m n=(\sec \theta+\tan \theta)(\sec \theta-\tan \theta) \\
& \Rightarrow \text { LHS }=\sec ^2 \theta-\tan ^2 \theta \quad\left[\text { Because }(a-b)(a+b)=a^2-b^2\right] \\
& \Rightarrow \text { LHS }=1\left[\text { Since } 1+\tan ^2 \theta=\sec ^2 \theta\right]
\end{aligned}
$
Hence, $m n=1$
View full question & answer→Question 152 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
$
Answerwe get:
$
\begin{aligned}
& x^2=(a \cos \theta)^2=a^2 \cos ^2 \theta \\
& y^2=(b \cot \theta)^2=b^2 \cot ^2 \theta \\
& \text { LHS }=\frac{a^2}{x^2}-\frac{b^2}{y^2}=\frac{a^2}{a^2 \cos ^2 \theta}-\frac{b^2}{b^2 \cot ^2 \theta}=\frac{1}{\cos ^2 \theta}-\frac{1}{\cot ^2 \theta} \\
& \Rightarrow \text { LHS }=\sec ^2 \theta-\tan ^2 \theta=1\left[\text { Since } 1+\tan ^2 \theta=\sec ^2 \theta\right]
\end{aligned}
$
View full question & answer→Question 162 Marks
Prove the following identity :
$
\sec ^4 A-\sec ^2 A=\frac{\sin ^2 A}{\cos ^4 A}
$
Answer$
\begin{aligned}
& \sec ^4 A-\sec ^2 A=\frac{1}{\cos ^4 A}-\frac{1}{\cos ^2 A} \\
& =\frac{1-\cos ^2 A}{\cos ^4 A} \\
& =\frac{\sin ^2 A}{\cos ^4 A} \quad\left[\because \sin ^2 A=1-\cos ^2 A\right]
\end{aligned}
$
View full question & answer→Question 172 Marks
Prove the following identity :
$
(1+\cot A)^2+(1-\cot A)^2=2 \operatorname{cosec}^2 A
$
Answer$
\begin{aligned}
& (1+\cot A)^2+(1-\cot A)^2 \\
& =1+\cot ^2 A+2 \cot A+1+\cot ^2 A-2 \cot A \\
& =2+2 \cot ^2 A=2\left(1+\cot ^2 A\right) \\
& =2 \operatorname{cosec} c^2 A
\end{aligned}
$
View full question & answer→Question 182 Marks
Prove the following identity :
$
\frac{1}{\sin A+\cos A}+\frac{1}{\sin A-\cos A}=\frac{2 \sin A}{1-2 \cos ^2 A}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{1}{\sin A+\cos A}+\frac{1}{\sin A-\cos A} \\
& =\frac{\sin A-\cos A+\sin A+\cos A}{\sin ^2 A-\cos ^2 A} \\
& =\frac{2 \sin A}{1-\cos ^2 A-\cos ^2 A}=\frac{2 \sin A}{1-2 \cos ^2 A}
\end{aligned}
$
View full question & answer→Question 192 Marks
Prove the following identity:
$
\sqrt{\operatorname{cosec}^2 q-1}=\cos q \operatorname{cosec} q
$
Answer$
\begin{aligned}
& \sqrt{\operatorname{cosec} 2-1}=\cos q \operatorname{cosec} q \\
& \sqrt{\operatorname{cosec} 2-1}=\sqrt{\cot ^2 q}\left(Q \operatorname{cosec}{ }^2 q-1=\cot ^2 q\right) \\
& =\cot q=\frac{\cos q}{\sin q}=\cos q \cdot \frac{1}{\sin q} \\
& =\cos q \text { cosecq }
\end{aligned}
$
View full question & answer→Question 202 Marks
Prove the following identity :
$
\sin ^4 A+\cos ^4 A=1-2 \sin ^2 A \cos ^2 A
$
Answer$
\begin{aligned}
& \sin ^4 A+\cos ^4 A=1-2 \sin ^2 A \cos ^2 A \\
& \Rightarrow \sin ^4 A+\cos ^4 A+2 \sin ^2 A \cos ^2 A=1 \\
& \text { LHS }=\left(\sin ^2 A+\cos ^2 A\right)^2 \\
& =1=\text { RHS }
\end{aligned}
$
View full question & answer→Question 212 Marks
Prove the following identity:
$
(\cos A+\sin A)^2+(\cos A-\sin A)^2=2
$
Answer$
\begin{aligned}
& \text { LHS }=(\cos A+\sin A)^2+(\cos A-\sin A)^2 \\
& =\cos ^2 A+\sin ^2 A+2 \cos A \cdot \sin A+\cos ^2 A+\sin ^2 A-2 \cos A \cdot \sin A \\
& =2\left(\cos ^2 A+\sin ^2 A\right)=2=\text { RHS }
\end{aligned}
$
View full question & answer→Question 222 Marks
Prove the following identity :
$
(\sec A-\cos A)(\sec A+\cos A)=\sin ^2 A+\tan ^2 A
$
Answer$
\begin{aligned}
& \text { LHS }=(\sec A-\cos A)(\sec A+\cos A) \\
& =\left(\sec ^2 A-\cos ^2 A\right)=1+\tan ^2 A-\left(1-\sin ^2 A\right) \\
& =\tan ^2 A+\sin ^2 A=\text { RHS }
\end{aligned}
$
View full question & answer→Question 232 Marks
Prove the following identity:
$
\tan ^2 A-\sin ^2 A=\tan ^2 A \cdot \sin ^2 A
$
Answer$
\begin{aligned}
& \text { LHS }=\tan ^2 A-\sin ^2 A \\
& =\frac{\sin ^2 A}{\cos ^2 A}-\sin ^2 A=\frac{\sin ^2 A\left(1-\cos ^2 A\right)}{\cos ^2 A} \\
& =\frac{\sin ^2 A}{\cos ^2 A} \cdot \sin ^2 A=\tan ^2 A \cdot \sin ^2 A=\text { RHS }
\end{aligned}
$
View full question & answer→Question 242 Marks
Prove the following identity:
$
\cos ^4 A-\sin ^4 A=2 \cos ^2 A-1
$
Answer$
\begin{aligned}
& \text { LHS }=\cos ^4 A-\sin ^4 A \\
& =\left(\cos ^2 A-\sin ^2 A\right)\left(\cos ^2 A+\sin ^2 A\right) \\
& =\left\{\cos ^2 A-\left(1-\cos ^2 A\right)\right\}=2 \cos ^2 A-1=\text { RHS }
\end{aligned}
$
View full question & answer→Question 252 Marks
Prove the following identity :
$
\sec ^2 A+\operatorname{cosec}{ }^2 A=\sec ^2 A \operatorname{cosec}{ }^2 A
$
Answer$
\begin{aligned}
& \text { LHS }=\sec ^2 A+\operatorname{cosec}{ }^2 A \\
& =\frac{1}{\cos ^2 A}+\frac{1}{\sin ^2 A}=\frac{\sin ^2 A+\cos ^2 A}{\cos ^2 A \cdot \sin ^2 A} \\
& =\frac{1}{\cos ^2 A \cdot \sin ^2 A}=\sec ^2 A \operatorname{cosec}{ }^2 A=\text { RHS }
\end{aligned}
$
View full question & answer→Question 262 Marks
Prove the following identity:
$
(1-\tan A)^2+(1+\tan A)^2=2 \sec ^2 A
$
Answer$
\begin{aligned}
& \text { LHS }=(1-\tan A)^2+(1+\tan A)^2 \\
& =1+\tan ^2 A-2 \tan A+1+\tan ^2 A+2 \tan A \\
& =2\left(1+\tan ^2 A\right)=2 \sec ^2 A=\text { RHS }
\end{aligned}
$
View full question & answer→Question 272 Marks
Prove the following identity:
$
\sin ^2 A \cos ^2 B-\cos ^2 A \sin ^2 B=\sin ^2 A-\sin ^2 B
$
Answer$
\begin{aligned}
& \text { LHS }=\sin ^2 A\left(1-\sin ^2 B\right)-\left(1-\sin ^2 A\right) \sin ^2 B \\
& =\sin ^2 A-\sin ^2 A \cdot \sin ^2 B-\sin ^2 B+\sin ^2 A \cdot \sin ^2 B \\
& =\sin ^2 A-\sin ^2 B=\text { RHS }
\end{aligned}
$
View full question & answer→Question 282 Marks
Prove the following identity:
$
\frac{1+\sin A}{1-\sin A}=\frac{\operatorname{cosec} A+1}{\operatorname{cosec} A-1}
$
Answer$
\begin{aligned}
& \text { LHS }=\frac{1+\sin A}{1-\sin A} \\
& \text { RHS }=\frac{\operatorname{cosec} A+1}{\operatorname{cosec} A-1}=\frac{\frac{1}{\sin A}+1}{\frac{1}{\sin A}-1} \\
& =\frac{1+\sin A}{1-\sin A}
\end{aligned}
$
View full question & answer→Question 292 Marks
Prove the following identity :
$
\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}
$
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}=\text { RHS }
\end{aligned}
$
View full question & answer→Question 302 Marks
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
Answer$
\begin{aligned}
& \text { LHS }=\operatorname{cosec} \theta(1+\cos \theta)(\operatorname{cosec} \theta-\cot \theta) \\
& =\frac{1}{\sin \theta}(1+\cos \theta)\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right) \\
& =\frac{(1+\cos \theta)}{\sin \theta}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\
& =\frac{1-\cos ^2 \theta}{\sin ^2 \theta}=\frac{\sin ^2 \theta}{\sin ^2 \theta}=1=\text { RHS }
\end{aligned}
$
View full question & answer→Question 312 Marks
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Answer$
\begin{aligned}
& \text { LHS }=\sec \mathrm{A}(1+\sin \mathrm{A})(\sec \mathrm{A}-\tan \mathrm{A}) \\
& =\frac{1}{\cos A}(1+\sin A)\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right) \\
& =\frac{(1+\sin A)}{\cos A}\left(\frac{1-\sin A}{\cos A}\right)=\frac{1-\sin ^2 A}{\cos ^2 A} \\
& =\left(\frac{\cos ^2 A}{\cos ^2 A}\right)=1=\text { RHS }
\end{aligned}
$
View full question & answer→Question 322 Marks
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Answer$
\begin{aligned}
& \text { LHS }=\sec A(1-\sin A)(\sec A+\tan A) \\
& =\frac{1}{\cos A}(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right) \\
& =\frac{(1-\sin A)}{\cos A}\left(\frac{1+\sin A}{\cos A}\right)=\left(\frac{1-\sin ^2 A}{\cos ^2 A}\right) \\
& =\left(\frac{\cos ^2 A}{\cos ^2 A}\right) \\
& =1=\text { RHS }
\end{aligned}
$
View full question & answer→Question 332 Marks
Prove the following identity:
$
\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)=\sec \theta+\operatorname{cosec} \theta
$
Answer$
\begin{aligned}
& \sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)=\sec \theta+\operatorname{cosec} \theta \\
& \text { LHS }=\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta) \\
& =\sin \theta\left(1+\frac{\sin \theta}{\cos \theta}\right)+\cos \theta\left(1+\frac{\cos \theta}{\sin \theta}\right) \\
& =\sin \theta\left(\frac{\cos \theta+\sin \theta}{\cos \theta}\right)+\cos \theta\left(\frac{\sin \theta+\cos \theta}{\sin \theta}\right) \\
& =\cos \theta+\sin \theta\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right) \\
& =\cos \theta+\sin \theta\left(\frac{1}{\sin \theta} \frac{1}{\cos \theta}\right)=\sec \theta+\operatorname{cosec} \theta
\end{aligned}
$
View full question & answer→Question 342 Marks
Prove the following identity :
tanA+cotA=secAcosecA
Answer$
\begin{aligned}
& \tan \mathrm{A}+\cot \mathrm{A}=\sec \mathrm{A} \operatorname{cosec} \mathrm{A} \\
& \text { Consider LHS }=\tan \mathrm{A}+\cot \mathrm{A} \\
& \tan \mathrm{A}+\cot \mathrm{A}=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin ^2 A+\cos ^2 A}{\sin A \cdot \cos A} \\
& \Rightarrow \tan A+\cot A=\frac{1}{\sin A \cdot \cos A}=\frac{1}{\sin A} \frac{1}{\cos A} \\
& \Rightarrow \tan \mathrm{A}+\cot \mathrm{A}=\operatorname{cosec} \mathrm{A} \cdot \sec \mathrm{A}=\mathrm{RHS}
\end{aligned}
$
View full question & answer→Question 352 Marks
Prove the following identity:
$
\left(1-\cos ^2 \theta\right) \sec ^2 \theta=\tan ^2 \theta
$
Answer$
\begin{aligned}
& \left(1-\cos ^2 \theta\right) \sec ^2 \theta=\tan ^2 \theta \\
& \text { Consider L.H.S }=\sin ^2 \theta \frac{1}{\cos ^2 \theta} \\
& =\tan ^2 \theta=\text { RHS }
\end{aligned}
$
View full question & answer→Question 362 Marks
Prove the following identity:
$
\left(1-\sin ^2 \theta\right) \sec ^2 \theta=1
$
Answer$
\begin{aligned}
& \left(1-\sin ^2 \theta\right) \sec ^2 \theta=1 \\
& \text { Consider L.H.S }=\cos ^2 \theta \sec ^2 \theta \\
& =\cos ^2 \theta \times \frac{1}{\cos ^2 \theta}=1 \\
& =\text { R.H.S }
\end{aligned}
$
Hence proved.
View full question & answer→