class 12 maths inverse trigonometric functions

Find the value of $\sin \left( {2{{\tan }^{ - 1}}\frac{1}{3}} \right) + \cos \left( {{{\tan }^{ - 1}}2\sqrt 2 } \right)$.

VAVidaara Admin Asked 8d ago 0 views 0 answers
📘 Inverse Trigonometric Functions NCERT,Ex.2.3,Q.8,Page.36 SA

Find the value of $\sin \left( {2{{\tan }^{ - 1}}\frac{1}{3}} \right) + \cos \left( {{{\tan }^{ - 1}}2\sqrt 2 } \right)$.

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

We have, $\sin \left( {2{{\tan }^{ - 1}}\frac{1}{3}} \right) + \cos \left( {{{\tan }^{ - 1}}2\sqrt 2 } \right)$

$= \sin \left[ {{{\sin }^{ - 1}}\left\{ {\frac{{2 \times \frac{1}{3}}}{{1 + {{\left( {\frac{1}{3}} \right)}^2}}}} \right\}} \right] + \cos \left( {{{\cos }^{ - 1}}\frac{1}{3}} \right)$

and $\left. {{{\tan }^{ - 1}}(2\sqrt 2 ) = {{\cos }^{ - 1}}\frac{1}{3}} \right]$

$= \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{{\frac{2}{3}}}{{1 + \frac{1}{9}}}} \right)} \right] + \frac{1}{3}$

$= \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{{2 \times 9}}{{3 \times 10}}} \right)} \right] + \frac{1}{3} = \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right] + \frac{1}{3}$

$= \frac{3}{5} + \frac{1}{3} = \frac{{9 + 5}}{{15}} = \frac{{14}}{{15}}$

View the full step-by-step solution page & related questions →

Community Answers (0)

Log in to post your own answer or join the discussion.

Discussion (0)

No comments yet — start the discussion.

← Back to all questions