4 Marks

Question 14 Marks

Prove that :\[\tan ^{-1}\left\{\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right\}=\frac{\pi}{4}+\frac{x}{2}, 0 < x <\frac{\pi}{2}\]

Answer

$
\begin{array}{l}
\tan ^{-1}\left\{\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right\} \\
=\tan ^{-1}\left\{\frac{\sqrt{2 \cos ^2 \frac{x}{2}}+\sqrt{2 \sin ^2 \frac{x}{2}}}{\sqrt{2 \cos ^2 \frac{x}{2}}-\sqrt{2 \sin ^2 \frac{x}{2}}}\right\}
\end{array}
$
$\begin{array}{l}=\tan ^{-1}\left\{\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right\} \\ {\left[\because 0 < \frac{x}{2} <\frac{\pi}{4} \Rightarrow \cos \frac{x}{2}>0, \sin \frac{x}{2}> 0\right]} \\ =\tan ^{-1}\left\{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right\} \\ =\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\} \\ =\frac{\pi}{4}+\frac{x}{2}\left[\because 0 < x <\frac{\pi}{2} \therefore \frac{\pi}{4}<\frac{\pi}{4}+\frac{x}{2}<\frac{\pi}{2}\right]\end{array}$

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Question 24 Marks

Prove that $2 \tan ^{-1}\left\{\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \frac{x}{2}\right\}=\cos ^{-1}\left(\frac{\beta+\alpha \cos x}{\alpha+\beta \cos x}\right)$

Answer

Suppose $\tan ^{-1}\left\{\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \frac{x}{2}\right\}=\theta$$
\begin{aligned}
\therefore \quad \tan \theta & =\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \cdot \frac{x}{2} \\
\text { now L.H.S. } & =2 \theta=\cos ^{-1}(\cos 2 \theta) \\
& =\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\left[\because \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right]
\end{aligned}
$
Putting the values :$
=\cos ^{-1}\left(\frac{1-\frac{\alpha-\beta}{\alpha+\beta} \tan ^2 \frac{x}{2}}{1+\frac{\alpha-\beta}{\alpha+\beta} \tan ^2 \frac{x}{2}}\right)
$
$\begin{array}{l}=\cos ^{-1}\left(\frac{(\alpha+\beta) \cos ^2 \frac{x}{2}-(\alpha-\beta) \sin ^2 \frac{x}{2}}{(\alpha+\beta) \cos ^2 \frac{x}{2}+(\alpha-\beta) \sin ^2 \frac{x}{2}}\right) \\ =\cos ^{-1}\left(\frac{\beta\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+\alpha\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}{\alpha\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+\beta\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}\right) \\ =\cos ^{-1}\left(\frac{\beta+\alpha \cos x}{\alpha+\beta \cos x}\right)=\text { R.H.S. }\end{array}$

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Question 34 Marks

Write in the simplest form $\cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right)$, where $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$.

Answer

To simplify $\cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right)$ we will express $\frac{3}{5} \cos x+\frac{4}{5} \sin x$ in consine form.
Suppose, $\frac{3}{5}=r \cos \theta$ and $\frac{4}{5}=r \sin \theta$
$r=\sqrt{\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2}=1$
and $\tan \theta=\frac{r \sin \theta}{r \cos \theta}=\frac{4}{3}$
$\Rightarrow \cos \theta=\frac{3}{5}$ and $\sin \theta=\frac{4}{5}$ where $\theta=\tan ^{-1}\left(\frac{4}{3}\right)$$
\begin{aligned}
\therefore \cos ^{-1}\left(\frac{3}{5}\right. & \left.\cos x+\frac{4}{5} \sin x\right) \\
& =\cos ^{-1}(\cos \theta \cos x+\sin \theta \sin x) \\
& =\cos ^{-1}\{\cos (x-\theta)\} \\
& =x-\theta \\
& =x-\tan ^{-1} \frac{4}{3} \text {}
\end{aligned}
$

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Question 44 Marks

If $x, y, z \in[-1,1]$ such that $\sin ^{-1} x+\sin ^{-1} y+$ $\sin ^{-1} z=\frac{-3 \pi}{2}$ then find the value of $x^2+y^2+z^2$.

Answer

We know that for $x \in[-1,1]$, minimum value of$
\begin{array}{l}
\sin ^{-1} x \text { is } \frac{-\pi}{2} \text {. } \\
\therefore \quad \sin ^{-1} x \geq \frac{-\pi}{2}, \sin ^{-1} y \geq \frac{-\pi}{2} \text { and } \sin ^{-1} z \geq \frac{-\pi}{2} \\
\forall x, y, z \in[-1,1] \\
\Rightarrow \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z \geq\left(\frac{-\pi}{2}\right)+\left(\frac{-\pi}{2}\right)+\left(\frac{-\pi}{2}\right) \\
\Rightarrow \quad \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z \geq \frac{-3 \pi}{2} \\
\therefore \quad \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{-3 \pi}{2} \\
\Rightarrow \quad \sin ^{-1} x=\frac{-\pi}{2}, \sin ^{-1} y=\frac{-\pi}{2}, \sin ^{-1} z=\frac{-\pi}{2} \\
\Rightarrow \quad x=y=z=-1
\end{array}
$
then $x^2+y^2+z^2=(-1)^2+(-1)^2+(-1)^2=3$

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Question 54 Marks

If $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$ then prove that$\frac{x^2}{a^2}-\frac{2 x y}{a b} \cos \alpha+\frac{y^2}{b^2}=\sin ^2 \alpha$

Answer

Given that $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$$
\begin{array}{l}
\cos ^{-1}\left\{\frac{x}{a} \cdot \frac{y}{b}-\sqrt{1-\frac{x^2}{a^2}} \sqrt{1-\frac{y^2}{b^2}}\right\}=\alpha \\
{\left[\because \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left\{x y-\sqrt{1-x^2} \sqrt{1-y^2}\right\}\right]} \\
\text { or } \quad \frac{x y}{a b}-\sqrt{1-\frac{x^2}{a^2}} \sqrt{1-\frac{y^2}{b^2}}=\cos \alpha \\
\quad\left(\frac{x y}{a b}-\cos \alpha\right)=\sqrt{1-\frac{x^2}{a^2}} \sqrt{1-\frac{y^2}{b^2}}
\end{array}
$
Square both sides
or $\quad\left(\frac{x y}{a b}-\cos \alpha\right)^2=\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)$
or $\frac{x^2 y^2}{a^2 b^2}-\frac{2 x y}{a b} \cos \alpha+\cos ^2 \alpha=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2 y^2}{a^2 b^2}$
or $\frac{x^2}{a^2}-\frac{2 x y}{a b} \cos \alpha+\frac{y^2}{b^2}=1-\cos ^2 \alpha$
or $\quad \frac{x^2}{a^2}-\frac{2 x y}{a b} \cos \alpha+\frac{y^2}{b^2}=\sin ^2 \alpha$ Hence proved.

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Question 64 Marks

Prove that $\frac{1}{2} \tan ^{-1} x=\cos ^{-1}\left\{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}\right\}^{\frac{1}{2}}$

Answer

$
\begin{array}{l}
\frac{1}{2} \tan ^{-1} x=\cos ^{-1}\left\{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}\right\}^{\frac{1}{2}} \\
\text { suppose } x=\tan \theta \\
\therefore \quad \text { L.H.S. }=\frac{1}{2} \tan ^{-1} x \\
=\frac{1}{2} \tan ^{-1}(\tan \theta)=\frac{1}{2} \times \theta=\frac{\theta}{2} \\
\because \tan ^{-1}(\tan \theta)=\theta
\end{array}
$hence L.H.S. $=\frac{\theta}{2}$$
\begin{array}{l}
\text { now } \quad \frac{\theta}{2}=\cos ^{-1}\left\{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}\right\}^{\frac{1}{2}} \\
\Rightarrow \quad \cos \frac{\theta}{2}=\left[\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}\right]^{\frac{1}{2}} \\
\text { R.H.S. }=\left[\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}\right]^{\frac{1}{2}}=\left[\frac{1+\sqrt{1+\tan ^2 \theta}}{2 \sqrt{1+\tan ^2 \theta}}\right]^{\frac{1}{2}} \\
\because x=\tan \theta
\end{array}
$
$\begin{array}{l}=\left[\frac{1+\sqrt{\sec ^2 \theta}}{2 \sqrt{\sec ^2 \theta}}\right]^{-\frac{1}{2}} \because 1+\tan ^2 \theta=\sec ^2 \theta \\ =\left[\frac{1+\sec \theta}{2 \sec \theta}\right]^{\frac{1}{2}}=\left[\frac{1}{2}\left(\frac{1}{\sec \theta}+1\right)\right]^{\frac{1}{2}} \\ =\sqrt{\frac{1+\cos \theta}{2}}=\sqrt{\frac{2 \cos ^2 \theta / 2}{2}} \\ =\cos \theta / 2=\text { L.H.S. Hence proved. }\end{array}$

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Question 74 Marks

Prove that $
\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{1+x^2}{2+x^2}}
$

Answer

$
\begin{array}{l}
\text { L.H.S. }=\cos \left[\tan ^{-1} \sin \left(\cot ^{-1} x\right)\right] \\
\text { Suppose } \quad \cot ^{-1} x=t \Rightarrow \cot t=x \\
\therefore \quad \operatorname{cosec}^2 t=1+\cot ^2 t \\
\operatorname{cosec} t=\sqrt{1+x^2}
\end{array}
$
$
\begin{array}{l}
\therefore \quad \sin t=\frac{1}{\sqrt{1+x^2}} \\
\therefore \quad \text { L.H.S. }=\cos \left[\tan ^{-1} \sin t\right] \\
=\cos \left[\tan ^{-1} \frac{1}{\sqrt{1+x^2}}\right] \\
\operatorname{suppose} \tan ^{-1} \frac{1}{\sqrt{1+x^2}}=Z \Rightarrow \tan Z=\frac{1}{\sqrt{1+x^2}} \\
\therefore \quad \sec ^2 Z=1+\tan ^2 Z \\
=1+\frac{1}{1+x^2}=\frac{1+x^2+1}{1+x^2} \\
=\frac{2+x^2}{1+x^2} \\
\therefore \quad \sec Z=\sqrt{\frac{2+x^2}{1+x^2}} \Rightarrow \cos Z=\sqrt{\frac{1+x^2}{2+x^2}} \\
\text { So, } \quad \text { L.H.S. }=\cos Z \\
=\sqrt{\frac{1+x^2}{2+x^2}}=\text { R.H.S. }
\end{array}
$

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