Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on trigonometric substitution. Click on this to access more such competitive quizzes on trigonometry.

Trigonometric substitution Quiz 2

mathematics

 Write the value of $\tan \left[ \dfrac{1}{2}{{\cos }^{-1}}\dfrac{\sqrt{7}}{3} \right]$

 Take value of $\tan \dfrac{M}{2}$ in terms of $\operatorname{cosM}$ and then substitute value of $\operatorname{cosM}$ in the same expression.

  $\text{Substitute } {{\cos }^{-1}}\dfrac{\sqrt{7}}{3}=M\\\Rightarrow \cos M=\dfrac{\sqrt{7}}{3}\\ \text{Use the formula, }\tan \dfrac{M}{2}=\sqrt{\dfrac{1-\cos M}{1+\operatorname{cosM}}}\\\therefore \tan \dfrac{M}{2}=\sqrt{\dfrac{1-\dfrac{\sqrt{7}}{3}}{1+\dfrac{\sqrt{7}}{3}}}=\sqrt{\dfrac{3-\sqrt{7}}{3+\sqrt{7}}}\\ \text{By rationalizing we get,}\\ \sqrt{\dfrac{3-\sqrt{7}}{3+\sqrt{7}}}=\sqrt{\dfrac{{{(3-\sqrt{7})}^{2}}}{9-7}}\\ \Rightarrow \dfrac{3-\sqrt{7}}{\sqrt{2}}$

 Write the value of ${{\sin }^{-1}}\left( \dfrac{2y}{1+{{y}^{2}}} \right)$ into the simplest form.

 Substitute $y=\tan \theta $ and then use half angle formula for $\tan \theta $ to further solve the expression.

  $\text{Consider the given trigonometric expression}\\ \text{Let }x={{\sin }^{-1}}\left( \dfrac{2y}{1+{{y}^{2}}} \right)\\\text{Substitute }y=\tan \theta \\\theta ={{\tan }^{-1}}y-----(i)\\ x={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)\\\Rightarrow {{\sin }^{-1}}\sin 2\theta \\\Rightarrow 2\theta \\\therefore x=2{{\tan }^{-1}}y\,\,\,\,\,\,\,\,\,(\text{Using equation (i)})$

 If $L=\tan^2 (co{{s}^{-1}}\theta )$, then find the value of $L$:

 Substitute ${{\cos }^{-1}}\theta $ as $y$ and then use the identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ to further simplify. 

  $\text{Substitute }{{\cos }^{-1}}\theta =y\\\Rightarrow \theta =\cos y-----(i)\\L=\tan y\\\Rightarrow {{\sec }^{2}}y-1\\=\dfrac{1}{{{\cos }^{2}}y}-1\\\Rightarrow \dfrac{1-{{\cos }^{2}}y}{{{\cos }^{2}}y}\\\Rightarrow \dfrac{1-{{\theta }^{2}}}{{{\theta }^{2}}}\,\,\,\,\,\,\,\,\,(\text{Using equation (i)})\\ \text{Substitute }\theta =\pi \\\therefore L=\dfrac{1-{{\pi }^{2}}}{{{\pi }^{2}}}$

 What is the value of $\sin \left( 2{{\tan }^{-1}}\dfrac{3}{8} \right)$?

 Use half angle formula of $\sin 2\theta $ in terms of $\tan \theta $ to simplify.

  $\text{Substitute }{{\tan }^{-1}}\dfrac{3}{8}=\theta \\\Rightarrow \tan \theta =\dfrac{3}{8}\\\sin \left( 2{{\tan }^{-1}}\dfrac{3}{8} \right)=\sin 2\theta \\\Rightarrow \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }\\\Rightarrow \dfrac{2\times \dfrac{3}{8}}{1+\dfrac{9}{64}}\Rightarrow \dfrac{\dfrac{3}{4}}{\dfrac{73}{64}}\\ \Rightarrow \dfrac{48}{73}$

 Write the value of ${{e}^{\sin \left[ {{\tan }^{-1}}\dfrac{7x}{2}+{{\cot }^{-1}}\dfrac{7x}{2} \right]}}$.

 Use property of inverse trigonometric functions to simplify the given expression. 

  $\text{Consider the given expression }\,\,\,{{e}^{\sin \left[ {{\tan }^{-1}}\dfrac{7x}{2}+{{\cot }^{-1}}\dfrac{7x}{2} \right]}}\\\Rightarrow {{e}^{\sin \pi /2}}, \,\,\,\,\,\,\,\,\left[\text{Use }{{\cot }^{-1}}\theta +{{\tan }^{-1}}\theta =\dfrac{\pi }{2}\right]\\\Rightarrow e$