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Question: If \({\cos ^{ - 1}}x = \alpha ,(0 < x < 1)\) and \({\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) + {\sec ^{ -...

If cos⁡−1x=α,(0<x<1){\cos ^{ - 1}}x = \alpha ,(0 < x < 1) and sin⁡−1(2x1−x2)+sec⁡−1(12x2−1)=2π3,{\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right) = \dfrac{{2\pi }}{3}, then tan⁡−1(2x){\tan ^{ - 1}}(2x) is equal to

A. π6 B. π4 C. π3 D. π2  A.{\text{ }}\dfrac{\pi }{6} \\\ B.{\text{ }}\dfrac{\pi }{4} \\\ C.{\text{ }}\dfrac{\pi }{3} \\\ D.{\text{ }}\dfrac{\pi }{2} \\\
Explanation

Solution

Hint- To evaluate the value of tan⁡−1(2x){\tan ^{ - 1}}(2x) we will first find the value of xx with the help of given equation, for it we will use some trigonometric formulas such as sin⁡2a=2sin⁡acos⁡a and cos⁡2a=2cos⁡2a−1\sin 2a = 2\sin a\cos a{\text{ and }}\cos 2a = 2{\cos ^2}a - 1

Complete step-by-step answer:
Given that, cos⁡−1x=α where (0<x<1)................(1){\cos ^{ - 1}}x = \alpha {\text{ where }}(0 < x < 1)................(1)
Therefore x=cos⁡αx = \cos \alpha
And given equation is sin⁡−1(2x1−x2)+sec⁡−1(12x2−1)=2π3{\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right) = \dfrac{{2\pi }}{3}
Now substitute the value of x=cos⁡αx = \cos \alpha in the above equation, we get
⇒sin⁡−1(2cos⁡α1−cos⁡2α)+sec⁡−1(12cos⁡2α−1)=2π3\Rightarrow {\sin ^{ - 1}}(2\cos \alpha \sqrt {1 - {{\cos }^2}\alpha } ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{2{{\cos }^2}\alpha - 1}}} \right) = \dfrac{{2\pi }}{3}
As we know that
1−cos⁡2A=sin⁡2A 2sin⁡Acos⁡A=sin⁡2A 2cos⁡2A−1=cos⁡2A  1 - {\cos ^2}A = {\sin ^2}A \\\ 2\sin A\cos A = \sin 2A \\\ 2{\cos ^2}A - 1 = \cos 2A \\\
Now, using the above formulas, we obtain
⇒sin⁡−1(2cos⁡α1−cos⁡2α)+sec⁡−1(12cos⁡2α−1)=2π3 ⇒sin⁡−1(2cos⁡αsin⁡2α)+sec⁡−1(1cos⁡2α)=2π3 ⇒sin⁡−1(sin⁡2α)+sec⁡−1(1cos⁡2α)=2π3 ⇒sin⁡−1(sin⁡2α)+sec⁡−1(sec⁡2α)=2π3 ⇒2α+2α=2π3 ⇒4α=2π3 ⇒α=π6  \Rightarrow {\sin ^{ - 1}}(2\cos \alpha \sqrt {1 - {{\cos }^2}\alpha } ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{2{{\cos }^2}\alpha - 1}}} \right) = \dfrac{{2\pi }}{3} \\\ \Rightarrow {\sin ^{ - 1}}(2\cos \alpha \sqrt {{{\sin }^2}\alpha } ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{\cos 2\alpha }}} \right) = \dfrac{{2\pi }}{3} \\\ \Rightarrow {\sin ^{ - 1}}(\sin 2\alpha ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{\cos 2\alpha }}} \right) = \dfrac{{2\pi }}{3} \\\ \Rightarrow {\sin ^{ - 1}}(\sin 2\alpha ) + {\sec ^{ - 1}}\left( {\sec 2\alpha } \right) = \dfrac{{2\pi }}{3} \\\ \Rightarrow 2\alpha + 2\alpha = \dfrac{{2\pi }}{3} \\\ \Rightarrow 4\alpha = \dfrac{{2\pi }}{3} \\\ \Rightarrow \alpha = \dfrac{\pi }{6} \\\
From equation (1)
∵x=cos⁡π6=32 ⇒2x=3  \because x = \cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2} \\\ \Rightarrow 2x = \sqrt 3 \\\
Therefore, the value of tan⁡−1(2x){\tan ^{ - 1}}(2x) is
tan⁡−1(2x)=tan⁡−1(3) =π3  {\tan ^{ - 1}}(2x) = {\tan ^{ - 1}}(\sqrt 3 ) \\\ = \dfrac{\pi }{3} \\\
Hence, the value of tan⁡−1(2x){\tan ^{ - 1}}(2x) is π3\dfrac{\pi }{3}

Note- To solve these types of questions, memorize all the formulas of trigonometry like allied angle, addition, double angle, triple angle etc. Understand the concept of domain and range. As in above question, the function is given as cos⁡−1x=α where (0<x<1){\cos ^{ - 1}}x = \alpha {\text{ where }}(0 < x < 1) and we make the function in terms of x such as x=cos⁡αx = \cos \alpha . So, in this type of questions try to convert inverse terms to solve the questions.