question_answer

  Prove that \[{{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x-a\sin x} \right)={{\tan }^{-1}}\left( \frac{a}{b} \right)-x.\] Or Prove that \[{{\cot }^{-1}}\left[ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right]=\frac{x}{2},\] \[x\in \left( 0,\,\,\frac{\pi }{4} \right).\]

Answer:

We have, \[LHS={{\tan }^{-\,1}}\left( \frac{a\cos x-b\sin x}{b\cos x-a\sin x} \right)={{\tan }^{-\,1}}\left\{ \frac{\frac{a\cos x-b\sin x}{b\cos x}}{\frac{b\cos x-a\sin x}{b\cos x}} \right\}\][on dividing numerator and denominator by \[b\cos x\]] \[={{\tan }^{-\,1}}\left\{ \frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x} \right\}={{\tan }^{-\,1}}\left( \frac{p-q}{1+pq} \right),\] where\[\frac{a}{b}=p\]and \[\tan x=q\] \[={{\tan }^{-\,1}}p-{{\tan }^{-\,1}}q={{\tan }^{-\,1}}\frac{a}{b}-{{\tan }^{-\,1}}(\tan x)\] \[=\left( {{\tan }^{-\,1}}\frac{a}{b}-x \right)=RHS\] Hence, \[{{\tan }^{-\,1}}=\left( \frac{a\cos x-b\sin x}{b\cos x-a\sin x} \right)=\left( {{\tan }^{-\,1}}\frac{a}{b}-x \right)\]. \[LHS={{\cot }^{-\,1}}\left[ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right]\] Consider,\[\sqrt{1+\sin x}\] \[=\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}\] \[={{\cot }^{-\,1}}\left[ \frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}} \right]={{\cot }^{-\,1}}\left[ \cot \frac{x}{2} \right]=\frac{x}{2}=RHS\] Similarly,\[\sqrt{1-\sin x}=\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}\] \[=\sqrt{{{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)}^{2}}}=\cos \frac{x}{2}-\sin \frac{x}{2}\] \[\left[ \because x\in \left[ 0,\,\,\frac{\pi }{4} \right],\sqrt{{{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)}^{2}}}=\cos \frac{x}{2}-\sin \frac{x}{2} \right]\] \[\therefore \]\[LHS={{\cot }^{-\,1}}\left[ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right]\] \[={{\cot }^{-\,1}}\left[ \frac{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)+\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)}{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)-\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)} \right]\]   \[={{\cot }^{-\,1}}\left[ \frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}} \right]={{\cot }^{-\,1}}\left[ \cot \frac{x}{2} \right]=\frac{x}{2}=RHS\] Hence proved.