Question 1 - Jee advanced Math 2022 P1 Questions with Solutions

Considering only the principal values of the inverse trigonometric function, the value of

\(\frac{3}{2} \cos^{-1}{\sqrt{\frac{2}{2+\pi^{2}}}} + \frac{1}{4}\sin^{-1}{\frac{2\sqrt{2}\pi}{2 + \pi^{2}}} + \tan^{-1}{\frac{\sqrt{2}}{\pi}}\)

is ____.


Sol :


Convert \(\cos^{-1}\) and \(\sin^{-1}\) terms into \(\tan^{-1}\).


To Convert \(\cos^{-1}\) term into \(\tan^{-1}\) :


Let \(t = \cos^{-1}{\sqrt{\frac{2}{2+\pi^{2}}}} = \cos^{-1}{\frac{\sqrt{2}}{\sqrt{2+\pi^{2}}}}\) …(1)


\(\implies\)\(\cos{t} = \frac{\sqrt{2}}{\sqrt{2+\pi^{2}}}\)



Use right angle triangle to determine ‘\(\tan{t}\)’. Consider a right triangle as shown above, with side adjacent to angle ‘t’ equal to \(\sqrt{2}\), and hypotenuse equal to \(\sqrt{2+\pi^{2}}\). 

Using Pythagoras’ theorem, 

side opposite to ‘t’ \(= \sqrt{2+\pi^{2} - 2} = \pi\)


Hence, 

\(\sin{t} = \frac{\pi}{\sqrt{2+\pi^{2}}}\)


And, 

\(\tan{t} = \frac{\sin{t}}{\cos{t}}\)

        \(  = \frac{\pi}{\sqrt{2}}\)


Therefore,

\(t = \tan^{-1}{\frac{\pi}{\sqrt{2}}}\)


So from (1), the first term can be re-written as \(\frac{3}{2}\tan^{-1}{\frac{\pi}{\sqrt{2}}}\).


To convert \(\sin^{-1}\) term into \(\tan^{-1}\) : 


We will do the same thing for the second term.


Let \(w = \sin^{-1}{\frac{2\sqrt{2}\pi}{2+\pi^{2}}}\)


\(\implies\)\(\sin{w} = \frac{2\sqrt{2}\pi}{2+\pi^{2}}\)


From the right triangle,


side adjacent to ‘w’ \(= \sqrt{(2+\pi)^{2} - 8\pi^{2}}\)

\( = \sqrt{4+4\pi^{2}+\pi^{4}-8\pi^{2}}\)

\( = \sqrt{\pi^{4}-4\pi^{2}+4}\)

\( = \sqrt{(\pi^{2}-2)^{2}}\)

\( = \pi^{2} - 2\)


Hence, 

\(\cos{w} = \frac{\pi^{2}-2}{\pi^{2}+2}\)


And,

\(\tan{w} = \frac{2\sqrt{2}\pi}{\pi^{2}-2}\)


Therefore,

\(w = \tan^{-1}{\frac{2\sqrt{2}\pi}{\pi^{2}-2}}\)


So the second term can be re-written as \(\frac{1}{4}\tan^{-1}{\frac{2\sqrt{2}\pi}{\pi^{2}-2}}\).


The entire expression can then be re-written as:


\(\frac{3}{2}\tan^{-1}{\frac{\pi}{\sqrt{2}}} + \frac{1}{4}\tan^{-1}{\frac{2\sqrt{2}\pi}{\pi^{2}-2}} + \tan^{-1}{\frac{\sqrt{2}}{\pi}}\)


Focus on the second term again. With a little bit of math trickery it can be further re-written as \(\frac{1}{4}\tan^{-1}{\frac{2(\frac{\pi}{\sqrt{2}})}{-(1 - (\frac{\pi}{\sqrt{2}})^2)}}\),


which equals \(-\frac{1}{4}\tan^{-1}{\frac{2(\frac{\pi}{\sqrt{2}})}{(1 - (\frac{\pi}{\sqrt{2}})^2)}}\).


By letting \(\tan{\theta} = \frac{\pi}{\sqrt{2}}\), and using the following double-angle formula :

\(\tan{2\theta} = \frac{2\tan{\theta}}{1 - \tan^{2}{\theta}}\)


the second term then becomes 

\(-\frac{1}{4}\tan^{-1}{\tan{2\theta}} = -\frac{1}{4}\tan^{-1}{(\tan{(2\tan^{-1}{\frac{\pi}{\sqrt{2}}})})}\)


There’s quite a bit in it. Firstly, notice we can cancel \(tan^{-1}\) and \(\tan{}\) in the expression \(\tan^{-1}{\tan{(\text{…something…})}}\) only if that something is in the principal value branch of inverse tangent function, i.e., in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\). 


In our case, that something is \(2\tan^{-1}{\frac{\pi}{\sqrt{2}}}\).


The value of \(\frac{\pi}{\sqrt{2}}\) is roughly about 2.2, which is between 1 and \(\sqrt{3}\).

\(1 < \frac{\pi}{\sqrt{2}} < \sqrt{3}\)


tan inverse is an increasing function for increasing values of x.

\(\implies \tan^{-1}{1} < \tan^{-1}{\frac{\pi}{\sqrt{2}}} < \tan^{-1}{\sqrt{3}}\)


\(\implies \frac{\pi}{4} < \tan^{-1}{\frac{\pi}{\sqrt{2}}} < \frac{\pi}{3}\)


\(\implies \frac{\pi}{2} < 2\tan^{-1}{\frac{\pi}{\sqrt{2}}} < \frac{2\pi}{3}\)


So, that something is clearly not in the principal value branch. Instead it is in the branch immediately to the right of it \((\frac{\pi}{2}, \frac{3\pi}{2})\) as seen in the graph above(highlighted in red). So we subtract \(\pi\) from it for the corresponding equivalent in the principal value branch. 

\(\implies -\frac{1}{4}\tan^{-1}{\tan{2\theta}} = -\frac{1}{4}\tan^{-1}{(\tan{(2\tan^{-1}{\frac{\pi}{\sqrt{2}}})})}\)

\(= -\frac{1}{4}\tan^{-1}{(\tan{(2\tan^{-1}{\frac{\pi}{\sqrt{2}}} - \pi)})}\)


And now we can cancel \(\tan^{-1}\) and \(\tan\).


\(= \frac{\pi}{4} - \frac{1}{2}\tan^{-1}{\frac{\pi}{\sqrt{2}}}\)


The entire expression can then be re-written as:


\(\frac{3}{2}\tan^{-1}{\frac{\pi}{\sqrt{2}}} + \frac{\pi}{4} - \frac{1}{2}\tan^{-1}{\frac{\pi}{\sqrt{2}}} + \tan^{-1}{\frac{\sqrt{2}}{\pi}}\)


\(= \frac{\pi}{4} + \tan^{-1}{\frac{\pi}{\sqrt{2}}} + \tan^{-1}{\frac{\sqrt{2}}{\pi}}\)


\(= \frac{\pi}{4} + \tan^{-1}{\frac{\pi}{\sqrt{2}}} + \cot^{-1}{\frac{\pi}{\sqrt{2}}}\)


\(= \frac{\pi}{4} + \frac{\pi}{2}\)

\(= \frac{3\pi}{4}\)

\(\approx 2.36\)










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