Question

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

Answer 1

We are given a trigonometric expression as ${{\tan }^{-1}}\left[ 2\sin \left( 2{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} \right) \right]$ . To solve this expression we need to solve the inner brackets first , i.e. simplify the expression $2{{\cos }^{-1}}\dfrac{\sqrt{3}}{2}$with the use of inverse trigonometric identities, and get a value of angle. Then, substitute the angle in the given trigonometric expression and find the solution.

Complete step-by-step solution:
We are given a trigonometric expression as${{\tan }^{-1}}\left[ 2\sin \left( 2{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} \right) \right]......(1)$
Let us assume that $2{{\cos }^{-1}}\dfrac{\sqrt{3}}{2}=\theta ......(2)$
So, using inverse trigonometric identity, we can write equation (2) as:

& \Rightarrow {{\cos }^{-1}}\dfrac{\sqrt{3}}{2}=\dfrac{\theta }{2} \\\ & \Rightarrow \dfrac{\sqrt{3}}{2}=\cos \dfrac{\theta }{2}......(3) \\\ \end{aligned}$$ Since, $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ So, equation (3) can be written as: $$\begin{aligned} & \Rightarrow \cos \dfrac{\pi }{6}=\cos \dfrac{\theta }{2} \\\ & \Rightarrow \dfrac{\theta }{2}=\dfrac{\pi }{6} \\\ & \Rightarrow \theta =\dfrac{\pi }{3}......(4) \\\ \end{aligned}$$ Now, substitute the value of $\theta $ in equation (1), we get: $$\begin{aligned} & \Rightarrow {{\tan }^{-1}}\left[ 2\sin \left( 2{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} \right) \right] \\\ & \Rightarrow {{\tan }^{-1}}\left[ 2\sin \left( \dfrac{\pi }{3} \right) \right]......(5) \\\ \end{aligned}$$ Since, $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$. So, equation (5) can be written as: $\begin{aligned} & \Rightarrow {{\tan }^{-1}}\left[ 2\times \dfrac{\sqrt{3}}{2} \right] \\\ & \Rightarrow {{\tan }^{-1}}\left[ \sqrt{3} \right]......(6) \\\ \end{aligned}$ Since, $\tan \dfrac{\pi }{3}=\sqrt{3}$ Therefore, we can write ${{\tan }^{-1}}\left[ \sqrt{3} \right]=\dfrac{\pi }{3}$ **Hence, the solution of ${{\tan }^{-1}}\left[ 2\sin \left( 2{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} \right) \right]$ is $$\dfrac{\pi }{3}$$.** **Note:** Be careful while assuming $\theta $. Always go with the value given in the inner-most bracket. It makes the solution easier. If we assume the whole expression as $\theta $, then using inverse trigonometric identities, we will end up with a somewhat similar expression again. Also, while applying inverse trigonometric identities, always check whether the value provided is negative or positive because the identities vary with sign. In case, if it is given $${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$$ instead of $${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$$, we have to write $\pi -{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$. Hence, there is a possibility that the final answer might be negative. Also, try to build equations in such a way that you can substitute the values of the standard angle to get a solution. As in this question, we took a standard value of $\dfrac{\pi }{3}$ and substitutes in the given expression to get a solution.