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Q)

# Find the value of $$\tan^{-1} \bigg[ 2\cos \bigg(2\sin^{-1}\frac{1}{2}\bigg) \bigg]$$

$\begin{array}{1 1} \frac{\pi}{4} \\ \frac{\pi}{2} \\ \frac{\pi}{8} \\ \pi \end{array}$

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A)
• $$sin\large\frac{\pi}{6}=\frac{1}{2}$$
• $$cos\large\frac{\pi}{3}=\frac{1}{2}$$
• $$tan^{-1}1=\large\frac{\pi}{4}$$
Given $tan^{-1} \bigg[ 2cos \bigg(2sin^{-1}\large\frac{1}{2}\bigg) \bigg]$:
Let $$x=sin^{-1}\large\frac{1}{2}$$. Then $$sinx= \large\frac{1}{2}\Rightarrow\:x=\frac{\pi}{6}$$
The given expression becomes $$tan^{-1} \bigg[ 2cos \bigg( 2.\large\frac{\pi}{6} \bigg) \bigg]$$ $$= tan^{-1} \bigg[ 2cos \bigg(\large\frac{\pi}{3} \bigg) \bigg]$$.
Given that $$cos\large\frac{\pi}{3}=\large\frac{1}{2}$$, the expression reduces to $tan^{-1} \bigg[ 2.\large\frac{1}{2} \bigg] = tan^{-1}1$
We know that $$tan^{-1}1 = \large\frac{\pi}{4}$$