Question

# What will be the value of the expression $\sin \left[ {2{{\cos }^{ - 1}}\cot \left( {2{{\tan }^{ - 1}}\dfrac{1}{2}} \right)} \right]$.

Hint: In this problem we have to find the value of the given expression, use the basic conversion techniques of various inverse trigonometric ratios into one another and simplify using the various inverse trigonometric identities, to reach the answer.

Given equation is
$\sin \left[ {2{{\cos }^{ - 1}}\cot \left( {2{{\tan }^{ - 1}}\dfrac{1}{2}} \right)} \right]$
As we know $2{\tan ^{ - 1}}A = {\tan ^{ - 1}}\dfrac{{2A}}{{1 - {A^2}}}$
So use this property in above equation we have,
$\Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\cot \left( {{{\tan }^{ - 1}}\dfrac{{2 \times \dfrac{1}{2}}}{{1 - {{\left( {\dfrac{1}{2}} \right)}^2}}}} \right)} \right]$
Now on simplifying we have,
$\Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\cot \left( {{{\tan }^{ - 1}}\dfrac{4}{3}} \right)} \right]$
Now as we know ${\tan ^{ - 1}}\theta = {\cot ^{ - 1}}\dfrac{1}{\theta }$ so, use this property in above equation we have,
$\Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\cot \left( {{{\cot }^{ - 1}}\dfrac{3}{4}} \right)} \right]$
Now cot and cot inverse is cancel out
$\Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\dfrac{3}{4}} \right]$
Now as we know ${\cos ^{ - 1}}A + {\cos ^{ - 1}}B = {\cos ^{ - 1}}\left( {AB - \sqrt {\left( {1 - {A^2}} \right)\left( {1 - {B^2}} \right)} } \right)$ so, use this property in above equation we have,
$\Rightarrow \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{3}{4}.\dfrac{3}{4} - \sqrt {\left( {1 - {{\left( {\dfrac{3}{4}} \right)}^2}} \right)\left( {1 - {{\left( {\dfrac{3}{4}} \right)}^2}} \right)} } \right)} \right]$
$\Rightarrow \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{9}{{16}} - \left( {1 - {{\left( {\dfrac{3}{4}} \right)}^2}} \right)} \right)} \right]$

$\Rightarrow \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{9}{{16}} - \left( {1 - \dfrac{9}{{16}}} \right)} \right)} \right] = \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{9}{{16}} - \dfrac{7}{{16}}} \right)} \right]$
$= \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{2}{{16}}} \right)} \right] = \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{8}} \right)} \right]$â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
Now let $y = {\cos ^{ - 1}}\left( {\dfrac{1}{8}} \right)$
$\Rightarrow \cos y = \dfrac{1}{8}$
Now as we know that ${\sin ^2}y + {\cos ^2}y = 1$
$\Rightarrow {\sin ^2}y + {\left( {\dfrac{1}{8}} \right)^2} = 1 \\ \Rightarrow {\sin ^2}y = 1 - \dfrac{1}{{64}} = \dfrac{{63}}{{64}} \\ \Rightarrow \sin y = \sqrt {\dfrac{{63}}{{64}}} = \sqrt {\dfrac{{7 \times 9}}{{8 \times 8}}} = \dfrac{3}{8}\sqrt 7 \\$
$\Rightarrow y = {\sin ^{ - 1}}\dfrac{{3\sqrt 7 }}{8}$
Now substitute this value in equation (1) we have
$\Rightarrow \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{8}} \right)} \right] = \sin \left[ {{{\sin }^{ - 1}}\dfrac{{3\sqrt 7 }}{8}} \right]$
Now the sin and sin inverse is canceled out.
$\Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\cot \left( {2{{\tan }^{ - 1}}\dfrac{1}{2}} \right)} \right] = \dfrac{{3\sqrt 7 }}{8}$
Hence option (a) is correct.

Note: Whenever we face such type of problems the basics is knowing about the simple inverse trigonometric identities like $2{\tan ^{ - 1}}A = {\tan ^{ - 1}}\dfrac{{2A}}{{1 - {A^2}}}$ and many others are being mentioned above. Such problems are more formula based thus having a good grasp over the identities will help in getting on the right track to reach the answer.