Question

Match the following.

Column I		Column II
(i)	$ta{n}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=$	(p)	$ta{n}^{-1}\frac{4}{3}-x$
(ii)	$co{s}^{-1}\left(\frac{3}{5}cosx+\frac{4}{5}sinx\right)$, where $x\in \left(-\frac{3\pi }{4},\frac{\pi }{4}\right)$ is equal to	(q)	$\sqrt{\frac{1+x^2}{2+x^2}}$
(iii)	$co{s}^{-1}\left(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right)$, where $x\in \left(0,\frac{\pi }{4}\right)$ is equal to	(r)	$\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}{x}^2$
(iv)	$cos\left(ta{n}^{-1}\left(sin\left(co{t}^{-1}x\right)\right)\right)=$	(s)	$\frac{x}{2}$

• A. (i) $\to$ (s), (ii) $\to$ (q), (iii) $\to$ (r), (iv) $\to$ (p)
• B. (i) $\to$ (r), (ii) $\to$ (s), (iii) $\to$ (q), (iv) $\to$ (p)
• C. (i) $\to$ (r), (ii) $\to$ (p), (iii) $\to$ (s), (iv) $\to$ (q)
• D. (i) $\to$ (p), (ii) $\to$ (r), (iii) $\to$ (q), (iv) $\to$ (s)

Answer 1

Answer: (C)

$\left(i\right)$ Put $x^2=cos2\theta$
$\Rightarrow \theta =\frac{1}{2}co{s}^{-1}{x}^2...\left(i\right)$
$\therefore ta{n}^{-1}\left(\frac{\sqrt{1+cos2\theta }+\sqrt{1-cos2\theta }}{\sqrt{1+cos2\theta }-\sqrt{1-cos2\theta }}\right)$
$=ta{n}^{-1}\left(\frac{\sqrt{2}cos\theta +\sqrt{2}sin\theta }{\sqrt{2}cos\theta -\sqrt{2}sin\theta }\right)$
$=ta{n}^{-1}\left(\frac{cos\theta +sin\theta }{cos\theta -sin\theta }\right)$
$=ta{n}^{-1}\left(\frac{1+tan\theta }{1-tan\theta }\right)$
$=ta{n}^{-1}\left[tan\left(\frac{\pi }{4}+\theta \right)\right]$
$=\frac{\pi }{4}+\theta =\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}{x}^2$ [Using (i)]
$\left(ii\right)$ Let $cos\alpha =\frac{3}{5}$
$\Rightarrow sin\alpha =\frac{4}{5},tan\alpha =\frac{4}{3}$
$\Rightarrow \alpha =ta{n}^{-1}\left(\frac{4}{3}\right)$
$\therefore co{s}^{-1}\left[\frac{3}{5}cosx+\frac{4}{5}sinx\right]$
$=co{s}^{-1}\left[cos\alpha \cdot cosx+sin\alpha \cdot sinx\right]$
$=co{s}^{-1}\left[cos\left(\alpha -x\right)\right]$
$=\alpha -x=ta{n}^{-1}\frac{4}{3}-x$
$\left(iii\right)co{t}^{-1}\left(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right)$
$=co{t}^{-1}\left(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\times \frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}+\sqrt{1-sinx}}\right)$

$=co{t}^{-1}\left(\frac{{\left(\sqrt{1+sinx}+\sqrt{1-sinx}\right)}^2}{\left(1+sinx\right)-\left(1-sinx\right)}\right)$
$=co{t}^{-1}\left(\frac{\left(1+sinx\right)+\left(1-sinx\right)+2\sqrt{\left(1+sinx\right)\left(1-sinx\right)}}{2sinx}\right)$
$=co{t}^{-1}\left(\frac{2+2\sqrt{1-si{n}^{2}x}}{2sinx}\right)$
$=co{t}^{-1}\left(\frac{1+|cosx|}{sinx}\right)$
$=co{t}^{-1}\left(\frac{1+cosx}{sinx}\right)$
$\left(\because x\in \left(0,\frac{\pi }{4}\right)\Rightarrow cosx>0\Rightarrow \left|cosx\right|=cosx\right)$
$=co{t}^{-1}\left(\frac{2co{s}^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}\right)$
$=co{t}^{-1}\left(cot\frac{x}{2}\right)=\frac{x}{2}$.
$\left(iv\right)$ Let $co{t}^{-1}x=y$
$\Rightarrow x=coty$.
$\therefore cosecy=\sqrt{1+co{t}^{2}y}=\sqrt{1+x^2}$
$\Rightarrow siny=\frac{1}{\sqrt{1+x^2}}$.
Let $ta{n}^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)=z$
$\Rightarrow tanz=\frac{1}{\sqrt{1+x^2}}$
$\therefore secz=\sqrt{1+ta{n}^{2}z}=\sqrt{1+\frac{1}{\sqrt{1+x^2}}}=\sqrt{\frac{2+x^2}{1+x^2}}$
$\Rightarrow cosz=\sqrt{\frac{1+x^2}{2+x^2}}$.
Now, $cos\left(ta{n}^{-1}\left(sin\left(co{t}^{-1}x\right)\right)\right)=cos\left(ta{n}^{-1}\left(siny\right)\right)$
$=cos\left(ta{n}^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right)$
$=cosz=\sqrt{\frac{1+x^2}{2+x^2}}$