Question

Match the following.

• 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)

(i) $\to$ (r), (ii) $\to$ (p), (iii) $\to$ (s), (iv) $\to$ (q)

Answer 2

$\left(i\right)$ Put $x^{2 }= cos2\theta$ $\Rightarrow \theta = \frac{1}{2} cos^{-1}\,x^{2}\quad...\left(i\right)$ $\therefore tan^{-1} \left(\frac{\sqrt{1+cos\,2\theta}+\sqrt{1-cos\,2\theta }}{\sqrt{1+cos\,2\theta }-\sqrt{1-cos\,2\theta }}\right)$ $= tan^{-1}\left(\frac{\sqrt{2}\,cos\,\theta +\sqrt{2}\,sin\,\theta }{\sqrt{2}\,cos\,\theta -\sqrt{2}\,sin\,\theta }\right)$ $= tan^{-1} \left(\frac{cos\,\theta +sin\,\theta }{cos\,\theta -sin\,\theta }\right)$ $= tan^{-1}\left(\frac{1+tan\,\theta }{1-tan\,\theta }\right)$ $= tan^{-1}\left[tan\left(\frac{\pi}{4}+\theta\right)\right]$ $= \frac{\pi }{4}+\theta = \frac{\pi }{4}+\frac{1}{2} cos^{-1}\,x^{2}\,\,\,\,$ [Using (i)] $\left(ii\right)$ Let $cos\,\alpha = \frac{3}{5}$ $\Rightarrow sin\,\alpha = \frac{4}{5}, tan\,\alpha = \frac{4}{3}$ $\Rightarrow \,\alpha = tan^{-1} \left(\frac{4}{3}\right)$ $\therefore cos^{-1}\left[\frac{3}{5}cos\,x + \frac{4}{5}sin\,x\right]$ $= cos^{-1}\left[cos\,\alpha\cdot cosx + sin\,\alpha \cdot sinx \right]$ $= cos^{-1}\left[cos\,\left(\alpha -x\right)\right]$ $= \alpha - x = tan^{-1} \frac{4}{3} - x$ $\left(iii\right) cot^{-1} \left(\frac{\sqrt{1+sin\,x} + \sqrt{1-sin\,x}}{\sqrt{1+sin\,x} - \sqrt{1-sin\,x}}\right)$ $= cot^{-1} \left(\frac{\sqrt{1+sin\,x} + \sqrt{1-sin\,x}}{\sqrt{1+sin\,x} - \sqrt{1-sin\,x}}\times \frac{\sqrt{1+sin\,x} + \sqrt{1-sin\,x}}{\sqrt{1+sin\,x} + \sqrt{1-sin\,x}}\right)$ $= cot^{-1} \left(\frac{\left(\sqrt{1+sin\,x} + \sqrt{1-sin\,x}\right)^{2}}{\left(1+sin\,x\right)- \left(1-sin\,x\right)}\right)$ $= cot^{-1}\left(\frac{\left(1+sin\,x\right)+\left(1-sin\,x\right)+2\sqrt{\left(1+sin\,x\right)\left(1-sin\,x\right)}}{2\,sin\,x}\right)$ $= cot^{-1}\left(\frac{2+2\sqrt{1-sin^{2}\,x}}{2\,sin\,x}\right)$ $= cot^{-1}\left(\frac{1+\left|cos\,x\right|}{sin\,x}\right)$ $= cot^{-1}\left(\frac{1+cos\,x}{sin\,x}\right)$ $\left(\because x\in \left(0, \frac{\pi}{4}\right)\Rightarrow cos\,x > 0 \Rightarrow \left|cos\,x\right|= cos\,x\right)$ $= cot^{-1}\left(\frac{2cos^{2} \frac{x}{2}}{2sin \frac{x}{2} cos \frac{x}{2}}\right)$ $= cot^{-1}\left(cot \frac{x}{2}\right) = \frac{x}{2}$. $\left(iv\right)$ Let $cot^{-1} \,x = y$ $\Rightarrow x = cot\,y$. $\therefore cosec\,y = \sqrt{1+cot^{2}\,y} = \sqrt{1+x^{2}}$ $\Rightarrow sin\,y = \frac{1}{\sqrt{1+x^{2}}}$. Let $tan^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right) = z$ $\Rightarrow tan \,z =\frac{1}{\sqrt{1+x^{2}}}$ $\therefore sec\,z = \sqrt{1+ tan^{2}\,z} = \sqrt{1+\frac{1}{\sqrt{1+x^{2}}}} = \sqrt{\frac{2+x^{2}}{1+x^{2}}}$ $\Rightarrow cos\,z = \sqrt{\frac{1+x^{2}}{2+x^{2}}}$. Now, $cos\left(tan^{-1} \left(sin \left(cot^{-1} \,x\right)\right)\right) = cos\left(tan^{-1} \left(sin \,y\right)\right)$ $= cos\left(tan^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)\right)$ $= cos\,z = \sqrt{\frac{1+x^{2}}{2+x^{2}}}$