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Question: The value of \( {\tan ^{ - 1}}(2\sin ({\sec ^{ - 1}}(2))) \) is: a) \( \dfrac{\pi }{6} \) b) \(...

The value of tan⁡−1(2sin⁡(sec⁡−1(2))){\tan ^{ - 1}}(2\sin ({\sec ^{ - 1}}(2))) is:
a) π6\dfrac{\pi }{6}
b) π4\dfrac{\pi }{4}
c) π3\dfrac{\pi }{3}
d) π2\dfrac{\pi }{2}

Explanation

Solution

Hint : We will be using the reciprocal of inverse secant function and the values of trigonometric function at different angles from the trigonometric table to solve the problem. The approach is simple: first we will simplify the innermost term and then we will proceed forward to the outermost term. More precisely, first we will find out what sec⁡−1(2){\sec ^{ - 1}}(2) is, then you can think about what to find out next.
Formula used:
From trigonometry we have the following formulas:

  1. sec⁡−1(x)=cos⁡−1(1x){\sec ^{ - 1}}(x) = {\cos ^{ - 1}}(\dfrac{1}{x})
  2. cos⁡−1(12)=π3{\cos ^{ - 1}}(\dfrac{1}{2}) = \dfrac{\pi }{3}
  3. sin⁡(π3)=32\sin (\dfrac{\pi }{3}) = \dfrac{{\sqrt 3 }}{2}
  4. tan⁡−1(3)={\tan ^{ - 1}}(\sqrt 3 ) =

Complete step-by-step answer :
The given trigonometric expression is:
tan⁡−1(2sin⁡(sec⁡−1(2))){\tan ^{ - 1}}(2\sin ({\sec ^{ - 1}}(2)))
By using the inverse trigonometric identity sec⁡−1(x)=cos⁡−1(1x){\sec ^{ - 1}}(x) = {\cos ^{ - 1}}(\dfrac{1}{x}) the above expression can be written as:
⇒tan⁡−1(2sin⁡(cos⁡−1(12)))\Rightarrow {\tan ^{ - 1}}(2\sin ({\cos ^{ - 1}}(\dfrac{1}{2})))
From the trigonometric angle table we have cos⁡−1(12)=π3{\cos ^{ - 1}}(\dfrac{1}{2}) = \dfrac{\pi }{3} so,
tan⁡−1(2sin⁡(cos⁡−1(12)))=tan⁡−1(2sin⁡(π3)){\tan ^{ - 1}}(2\sin ({\cos ^{ - 1}}(\dfrac{1}{2}))) = {\tan ^{ - 1}}(2\sin (\dfrac{\pi }{3}))
Also using the trigonometric angle table we can write the above expression as:
⇒tan⁡−1(2×32)=tan⁡−1(3)\Rightarrow {\tan ^{ - 1}}(2 \times \dfrac{{\sqrt 3 }}{2}) = {\tan ^{ - 1}}(\sqrt 3 )
By using the trigonometric table again we can write the above expression as:
⇒tan⁡−1(3)=π3\Rightarrow {\tan ^{ - 1}}(\sqrt 3 ) = \dfrac{\pi }{3}
Therefore, the value of the expression tan⁡−1(2sin⁡(sec⁡−1(2))){\tan ^{ - 1}}(2\sin ({\sec ^{ - 1}}(2))) is π3\dfrac{\pi }{3}
So, the correct answer is “Option C”.

Note : You cannot solve the problem without simplifying it from the root. While solving, try to first deal with the angle part. Also use the trigonometric angle table wisely. The inverse of a trigonometric function always gives an angle while the general trigonometric function takes an angle to give a value. If you know the elementary sine, cosine, tangent function values at different angles and the identities related to them, then this kind of problem can be solved without having a peek into the trigonometric angle table.