question_answer

If \[x,\,\,y,\,\,z\in [-\,1,\,\,1],\] such that

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

OR

Prove that

\[2{{\tan }^{-1}}\left( \frac{1}{5} \right)+{{\sec }^{-1}}\left( \frac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \frac{1}{8} \right)=\frac{\pi }{4}.\]

Answer:

We know that, \[-\frac{\pi }{2}\le {{\sin }^{-1}}x\le \frac{\pi }{2},\] \[x\in [-\,1,\,\,1]\]

\[\therefore \] \[{{\sin }^{-1}}x\ge -\frac{\pi }{2},\] \[{{\sin }^{-1}}y\ge -\frac{\pi }{2},\] \[{{\sin }^{-1}}z\ge \frac{-\,\pi }{2}\] for all \[x,\text{ }y,\text{ }z\in [-\,1,\,\,1]\]

\[\Rightarrow \] \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z\]

\[\ge \left( -\,\frac{\pi }{2} \right)+\left( -\,\frac{\pi }{2} \right)+\left( -\,\frac{\pi }{2} \right)\]

\[\Rightarrow \]   \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z\ge \frac{-\,3\pi }{2}\]

But       \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\frac{-\,3\pi }{2}\]

\[\therefore \] \[{{\sin }^{-1}}x=-\frac{\pi }{2},\] \[{{\sin }^{-1}}y=-\frac{\pi }{2},\] \[{{\sin }^{-1}}z=-\frac{\pi }{2}\]

\[\Rightarrow \]   \[x=y=z=-\,1\]

Now, \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{(-\,1)}^{2}}+{{(-\,1)}^{2}}+{{(-\,1)}^{2}}=3\]

OR

\[LHS=2{{\tan }^{-1}}\left( \frac{1}{5} \right)+{{\sec }^{-1}}\left( \frac{5\sqrt{2}}{7} \right)+2\,\,{{\tan }^{-1}}\left( \frac{1}{8} \right)\]

\[=2\left( {{\tan }^{-1}}\frac{1}{5}+{{\tan }^{-1}}\frac{1}{8} \right)+{{\sec }^{-1}}\left( \frac{5\sqrt{2}}{7} \right)\]

\[=2\,\,{{\tan }^{-1}}\left[ \frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\times \frac{1}{8}} \right]+{{\tan }^{-1}}\sqrt{{{\left( \frac{5\sqrt{2}}{7} \right)}^{2}}-1}\]

If put \[{{\sec }^{-1}}x=\theta \] \[\Rightarrow \] \[x=\sec \theta \]

\[\Rightarrow \]   \[{{\sec }^{-1}}x={{\tan }^{-1}}\sqrt{{{x}^{2}}-1}]\]

Then, \[\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}\]

\[\Rightarrow \]   \[\theta ={{\tan }^{-1}}\sqrt{{{x}^{2}}-1}\]

\[=2\,\,{{\tan }^{-1}}\frac{13}{39}+{{\tan }^{-1}}\sqrt{\frac{50}{49}}-1\]

\[\therefore \] \[LHS=2{{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{7}={{\tan }^{-1}}\left\{ \frac{2\times \frac{1}{3}}{1-{{\left( \frac{1}{3} \right)}^{2}}} \right\}\]

\[+{{\tan }^{-1}}\frac{1}{7}\]

\[\left[ \because \,\,\,2{{\tan }^{-1}}x={{\tan }^{-1}}\left\{ \frac{2x}{1-{{x}^{2}}} \right\} \right]\]

\[={{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{1}{7}={{\tan }^{-1}}\left( \frac{\frac{3}{4}+\frac{1}{7}}{1-\left( \frac{3}{4}\times \frac{1}{7} \right)} \right)\]\[={{\tan }^{-1}}\left( \frac{21+4}{28-3} \right)={{\tan }^{-1}}(1)=ta{{n}^{-1}}\left( \tan \frac{\pi }{4} \right)\]          \[=\frac{\pi }{4}=RHS\]      \[[\because \,\,ta{{n}^{-1}}(tan\theta )=\theta ]\]

Hence proved.