# If ${\cos ^{ - 1}}x = \alpha ,(0 < x < 1)$ and ${\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right) = \dfrac{{2\pi }}{3},$ then ${\tan ^{ - 1}}(2x)$ is equal to

\[

A.{\text{ }}\dfrac{\pi }{6} \\

B.{\text{ }}\dfrac{\pi }{4} \\

C.{\text{ }}\dfrac{\pi }{3} \\

D.{\text{ }}\dfrac{\pi }{2} \\

\]

Last updated date: 14th Mar 2023

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Answer

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

Complete step-by-step answer:

Given that, ${\cos ^{ - 1}}x = \alpha {\text{ where }}(0 < x < 1)................(1)$

Therefore $x = \cos \alpha $

And given equation is ${\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right) = \dfrac{{2\pi }}{3}$

Now substitute the value of $x = \cos \alpha $ in the above equation, we get

$ \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 ^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

$

\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)

$

\because x = \cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2} \\

\Rightarrow 2x = \sqrt 3 \\

$

Therefore, the value of ${\tan ^{ - 1}}(2x)$ is

$

{\tan ^{ - 1}}(2x) = {\tan ^{ - 1}}(\sqrt 3 ) \\

= \dfrac{\pi }{3} \\

$

Hence, the value of ${\tan ^{ - 1}}(2x)$ is $\dfrac{\pi }{3}$

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 ${\cos ^{ - 1}}x = \alpha {\text{ where }}(0 < x < 1)$ and we make the function in terms of x such as $x = \cos \alpha $ . So, in this type of questions try to convert inverse terms to solve the questions.

Complete step-by-step answer:

Given that, ${\cos ^{ - 1}}x = \alpha {\text{ where }}(0 < x < 1)................(1)$

Therefore $x = \cos \alpha $

And given equation is ${\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) + {\sec ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right) = \dfrac{{2\pi }}{3}$

Now substitute the value of $x = \cos \alpha $ in the above equation, we get

$ \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 ^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

$

\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)

$

\because x = \cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2} \\

\Rightarrow 2x = \sqrt 3 \\

$

Therefore, the value of ${\tan ^{ - 1}}(2x)$ is

$

{\tan ^{ - 1}}(2x) = {\tan ^{ - 1}}(\sqrt 3 ) \\

= \dfrac{\pi }{3} \\

$

Hence, the value of ${\tan ^{ - 1}}(2x)$ is $\dfrac{\pi }{3}$

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 ${\cos ^{ - 1}}x = \alpha {\text{ where }}(0 < x < 1)$ and we make the function in terms of x such as $x = \cos \alpha $ . So, in this type of questions try to convert inverse terms to solve the questions.

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