Exercise 16 A
Question 1
Ans: $\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}=1$
$L \cdot H \cdot S=\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}$ $\left[\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right]$
$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta}=1=R \cdot \mathrm{H} \cdot \mathrm{S}$
Question 2
Ans: $\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta}$=1
L.H.S $=\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta} \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$
$=\frac{\cos ^{2} \theta}{\cos ^{2} \theta}=1=R \cdot H \cdot S$
Question 3
Ans: $\sin A \cdot \cot A=\cos A$
L.H.S=$\sin A \cdot \cot A$
$=\sin A \cdot \frac{\cos A}{\sin A}$ $\left[\cot A=\frac{\cos A}{\sin A}\right]$
$=\cos A=R . H .S$
Question 4
Ans: $\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta=1$
$L . H . S=\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta \quad\left[\frac{1}{\cos ^{2} \theta}=\sec ^{2} \theta\right]$
$=\sec ^{2} \theta-\tan ^{2} \theta$
$=1=R \cdot H \cdot S$
Question 5
Ans: $\tan ^{2} \cos ^{2} A=1-\cos ^{2} A$
$L \cdot H \cdot S=\tan ^{2} A \cos ^{2} A$ $\left[\tan ^{2} A=\frac{\sin ^{2} A}{\cos ^{2} A}\right]$
$=\frac{\sin ^{2} A}{\cos ^{2} A} \cos ^{2} A$
$=\sin ^{2} A=1-\cos ^{2} A=R \cdot H \cdot S$
Question 6
Ans: $\tan \theta=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$
$\Rightarrow R . H . S=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$ $\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$
$=\frac{\sin \theta}{\sqrt{\cos ^{2} \theta}}$
$=\frac{\sin \theta}{\cos \theta}=\tan \theta=$ L.H.S
Question 7
Ans: $\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1}{1-\cos \theta}$
L.H.S $=\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1+\cos \theta}{1-\cos ^{2} \theta}\left[\sin ^{2} \theta=1-\cos ^{2} \theta\right]$
$=\frac{1+\cos \theta}{(1-\cos \theta)(1-\cos \theta)}\left[a^{2}-b^{2}=(a-b)\left(a+
b\right)\right.$
b\right)\right.$
$=\frac{1}{1-\cos \theta}=$ R.H.S
Question 8
Ans: $\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)=\cos ^{2} \theta \quad\left[\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$L \cdot H \cdot S=\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)$ $\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right]$
$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \times \sin ^{2} \theta$
$=\cos ^{2} \theta$ = R.H.S
Question 9
Ans: $\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta$
L.H.S $=\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)$
$\begin{array}{ll}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \cos ^{2} \theta & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]}\end{array}$
$=\sin ^{2} \theta=R \cdot H \cdot S$
Question 10
Ans: $\left(1-\sin ^{2} \theta\right) \operatorname{sic}^{2} \theta=1$
L.H.S $=\left(1-\sin ^{2} \theta\right) \sec ^{2} \theta \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$
$=\cos ^{2} \theta \cdot \sec ^{2} \theta$
$=1=$ R.H.S
Question 11
Ans:
$\begin{aligned}\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta &=1 \\ L \cdot H \cdot S=\left(1-\cos ^{2} \theta\right) & \operatorname{cosec}^{2} \theta \\ 1-\cos ^{2} \theta &=\sin ^{2} \theta \\ \operatorname{cosec} \theta &=\frac{1}{\sin \theta} \\ &=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta} \\ 1^{-} &=R \cdot H \cdot S \end{aligned}$
Question 12
Ans: $\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=1$
$L \cdot H \cdot S=\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}$
$=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta} \quad\left[1+\tan ^{2} \theta=\sec ^{2} \theta\right]$
$\doteq \sin ^{2} \theta+\cos ^{2} \theta$ $\left[\frac{1}{\sec ^{2} \theta}=\cos ^{2} \theta\right]$
$R \cdot H .S=1$
Question 13
Ans: $\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}=1$
$L \cdot H S=\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}$
$\left[1+\cot ^{2} \theta=\cos \operatorname{sc}^{2} \theta\right]$
$=\cos ^{2} \theta+\frac{1}{\operatorname{cosec}^{2} \theta}$
$\left\{\frac{1}{\operatorname{cosec}}=\sin \theta\right.$
$=\cos ^{2} \theta+\sin ^{2} \theta$
$R \cdot H \cdot S=1$
Question 14
Ans: $\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$
$L \cdot H \cdot S=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$
$\sin ^{2} \theta+\cos ^{2} \theta=1$
$\sec ^{2} \theta-\tan ^{2} \theta=1$
$R \cdot H \cdot S=1$
Question 15
Ans: $\left[\frac{\cos ^{2} A}{\sin ^{2} A}+1\right] \tan ^{2} A=\frac{1}{\cos ^{2} A}$
$L \cdot H \cdot S=\left[\frac{\cos ^{2} \cdot A}{\sin ^{2} A}+1\right] \tan ^{2} \theta$
$\frac{\cos A}{\sin A}=\cot A$
$=\left(\cot ^{2} A+1\right) \tan ^{2} \theta$
$\quad \cos ^{2} A+1=\operatorname{cosec}^{2} A$
$=\operatorname{cosec}^{2} A \times \tan ^{2} A$
$\quad \operatorname{cosec} A=\frac{1}{\sin A}$
$=\frac{1}{\sin ^{2} A} \times \frac{\sin ^{2} A}{\cos ^{2} A}$
$R \cdot H \cdot S=\frac{1}{\cos ^{2} A}$
Question 16
Ans: $\sin ^{2} \theta+\sin ^{2} \theta \cos ^{2} \theta=\sin ^{2} \theta .$
$\begin{aligned} L \cdot H \cdot S=& \sin ^{4} \theta+\sin ^{2} \theta \cos ^{2} \theta \\=& \sin ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right.\\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=& \sin ^{2} \theta \times 1 \\ R \cdot H \cdot S=& \sin ^{2} \theta \end{aligned}$
Question 17
Ans: $\sin ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta+\cos 4 \theta=1$
$\begin{aligned} \text { L.H.S }=& \sin ^{2} \theta+2 \sin ^{2} \theta(a)^{2} \theta+\cos ^{4} \theta \\ & a^{2}+2 a b+b^{2}=(a+b)^{2} \\=&\left(\sin ^{2} \theta\right)^{2}+2 \sin ^{2} \theta \cos ^{2} \theta+\left(\cos ^{2} \theta\right)^{2} \\=&\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2} \\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=&(1)^{2} \end{aligned}$
R.H.S= 1
Question 18
Ans: $\sin 4 A \operatorname{cosec}^{2} A+\cos 4 A \sec ^{2} A=1$
L.H.S $=\sin 4 A \operatorname{cosec}^{2} A+\cos ^{4} A \sec ^{2} A$
$=\sin 4 A \times \frac{1}{\sin ^{2} A}+\cos ^{4} A \times \frac{1}{\cos ^{2} A}$ $\left[\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$
$\therefore \sin ^{2} \theta+\cos ^{2} \theta=1$ $\left[\sec \theta=\frac{1}{\cos \theta}\right]$
R.H.S =1
Question 19
Ans: $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$
$\begin{aligned} L \cdot H \cdot S=& \sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A \\=& \sin ^{2} A \times \frac{\cos ^{2} A}{\sin ^{2} A}+\cos ^{2} A \times \frac{\sin ^{2} A}{\cos ^{2} A} \\ & \frac{\cos A}{\sin A}=\cot \theta, \frac{\sin \theta}{\cos \theta}=\tan \theta \\=& \cos ^{2} A+\sin ^{2} A \cdot=1 \\ \text { R.H.S }=& 1 \end{aligned}$
Question 20
Ans: $\tan \theta+\cot \theta=\sec \theta \cdot \operatorname{cosec} \theta \cdot .$
$L \cdot H \cdot S=\tan \theta+\cot \theta$
${\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right] }$
$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}$
$=\left[\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta}\right]$
$=\frac{1}{\cos \theta \sin \theta}$
${\left[\frac{1}{\cos \theta}=\sec \theta, \frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }$
$R \cdot H \cdot S=\operatorname{Sec} \theta \cdot \operatorname{cosec} \theta$
Question 21
Ans: $(\tan A+\cot A) \sin A \cos A=1$
$L \cdot H \cdot S=(\tan A+\cot A)(\sin A \cos A)$
$\begin{aligned} & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta \doteq \frac{\cos \theta}{\sin \theta}\right] } \\=& {\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right] \sin A \cos A . } \\=& \frac{\sin ^{2} A+\cos ^{2} A}{\cos A \sin A} \times \sin A \cos A \\ & {\left[\sin ^{2} \theta+\operatorname{Cos}^{2} \theta=1\right] }\end{aligned}$
$=\frac{1}{\cos A \sin A} \times \sin A \cos A$
R.H.S = 1
Question 22
Ans: $\begin{aligned} \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta & \\ L \cdot H \cdot S=& \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)} \\=& \frac{1-\sin ^{2} \theta+\cos \theta}{\sin \theta(1+\cos \theta)} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{\cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta} \end{aligned}$
$=\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$
$=\frac{\cos \theta}{\sin \theta}$
${\left[\frac{\cos \theta}{\sin \theta}=\cot \theta\right] }$
$R \cdot H \cdot S=\cot \theta$
Question 23
Ans:
$\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=2 \operatorname{cosec}^{2} \theta$
$\begin{aligned} \text { L.H.S } &=\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta} \\ &=(a+b)(a-b)=a^{2}-b^{2} \\ &=\frac{1+\cos \theta+1-\cos \theta}{(1-\cos \theta)(1+\cos \theta)}=\frac{2}{1-\cos ^{2} \theta} \end{aligned}$
$\begin{aligned} & {\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] } \\=& \frac{2}{\sin ^{2} \theta} . \\ & {\left[\frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }\end{aligned}$
$R \cdot H \cdot S=2 \operatorname{cosec}^{2} \theta$
Question 24
Ans: $\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\tan ^{2} \theta .$
$L \cdot H \cdot S=\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\frac{1-\tan ^{2} \theta}{\frac{1}{\tan ^{2} \theta}-1}$
$=\frac{1-\tan ^{2} θ}{\frac{1-\tan ^{2} θ }{\tan ^{2} \theta}}$
$=\tan ^{2}\theta=R.H.S$
Question 25
Ans: $\frac{1}{\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$
$L \cdot H \cdot S=\frac{1}{\operatorname{Sec} \theta+\tan \theta}$
$=\frac{\sec \theta-\tan \theta}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$
$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta} \quad\left[\sec ^{2} \theta-\tan ^{2} \theta=1\right]$
$=\frac{\sec \theta-\tan \theta}{1}$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
R.H.S=$ \frac{1-\sin \theta}{\cos \theta}$
Question 26
Ans: $(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)=1$
L.H.S $=(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)$
$\left[\operatorname{cosec} A=\frac{1}{\sin A}, \sec A=\frac{1}{\cos A}, \tan A=\frac{\sin A}{\cos A}\right.$
$\left.\cot A=\frac{\cos A}{\sin A}\right]$
$=\left[\frac{1}{\sin A}-\sin A\right]\left[\frac{1}{\cos A}-\cos A\right]\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right]$
$=\frac{1-\sin ^{2} A}{\sin A} \times \frac{1-\cos ^{2} A}{\cos A} \times \frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$
$=\frac{\left[\sin ^{2} A+\cos ^{2} A=1\right]}{\sin A} \times \frac{\sin ^{2} A}{\cos A} \times \frac{1}{\sin A \cos A}$
$=\frac{\cos ^{2} A \sin ^{2} A}{\sin ^{2} A \cos ^{2} A}$
R.H.S $=1$
Question 27
Ans: $\frac{1+\sin \theta}{1-\sin \theta}=(\sec \theta+\tan \theta)^{2}$
$\begin{aligned} \text { L.H.S. } &=\frac{1+\sin \theta}{1-\sin \theta} \\ &=\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)} \\ &=\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta} \end{aligned}$
$\begin{aligned} & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta} \\=& {\left[\frac{1+\sin \theta}{\cos \theta}\right]^{2} } \\=& {\left[\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right]^{2} }\end{aligned}$
$\frac{1}{\cos \theta}=\sec \theta, \frac{\sin \theta}{\operatorname{con} \theta}=\tan \theta$
$R \cdot H \cdot S=(\sec \theta+\tan \theta)^{2}$
Question 28
Ans: $\frac{1+\cos \theta}{1-\cos \theta}=(\operatorname{cosec} \theta+\cot \theta)^{2}$
$\begin{aligned} L \cdot H \cdot S &=\frac{1+\cos \theta}{1-\cos \theta} \\ &=\frac{(1+\cos \theta)(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)} \cdot\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] \end{aligned}$
$=\frac{\left(1+\cos (\theta)^{2}\right.}{1-\cos ^{2} \theta}$
$=\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}$
$=\left[\frac{1+\cos \theta}{\sin \theta}\right]^{2}$
$=\left[\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right]^{2}$
$\left(\frac{1}{\sin \theta}=\operatorname{cosec} \theta, \frac{\cos \theta}{\sin \theta}=\cot \theta\right)$
$R \cdot H \cdot S=(\cos \theta+\theta+\cot \theta)^{2}$
Question 29
Ans: $\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\operatorname{cosec} \theta+\cot \theta-\frac{1+\cos \theta}{\sin \theta}$
$L \cdot H \cdot S=\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{\cot A+\operatorname{cosec} \theta-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{\cot \theta+\operatorname{cosec} \theta+\left(\cot ^{2} \theta-\operatorname{cosec} 2 \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)+\cot \theta+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\operatorname{cotc}+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta+1)}{(\cot \theta-\operatorname{cosec} \theta+1}$
$=\cot \theta+\operatorname{cosec} \theta$
$=\operatorname{cosec} \theta+\cot \theta$
$=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta} .$
$R \cdot H \cdot S=\frac{1+\cos \theta}{\sin \theta}$
Question 30
Ans: $\tan \theta+\cot \theta=2$
$(\tan \theta+\cot \theta)^{2}=4 \quad$ [Square both side]
$\tan ^{2} \theta+\cot ^{2} \theta+2 \tan \theta \cot \theta=4$
${[\tan \theta \cdot \cot \theta=1] }$
$\begin{aligned} \tan ^{2} \theta+\cot ^{2} \theta+2 \times 1 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta+2 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta &=4-2 \\ &=2 \\ \tan ^{2} \theta+\cot ^{2} \theta &=2 \end{aligned}$
Question 31
Ans: $a^{2}+b^{2}=(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2}$
$=-\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \epsilon \cos \theta$
$=2 \sin ^{2} \theta+2 \cos ^{2} \theta$
$=2\left(\sin ^{2} \theta+\left(\cos ^{2} \theta\right)\right.$
$=2 \times 1$
$=2$
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