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# If $2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\operatorname{co} {\text{sec }}x} \right)$ , then $x =$ ?(A) $\dfrac{\pi }{4}$ (B) $\dfrac{\pi }{6}$ (C) $\dfrac{\pi }{2}$ (D) No solution

Last updated date: 23rd Jun 2024
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Hint: Assume the given expression $2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\operatorname{co} {\text{sec }}x} \right)$ as some variable ‘m’. Now express the value of $m{\text{ and }}\dfrac{m}{2}$ as inverse tangent functions. On applying tangent on both sides, you will get the value for $\tan m{\text{ and tan}}\dfrac{m}{2}$ . Use the half-angle formula of a tangent, i.e. $\tan m = \dfrac{{2\tan \dfrac{m}{2}}}{{1 - {{\tan }^2}\dfrac{m}{2}}}$ and substitute the values. Now solve further to obtain the value of $\tan x$ and hence the value of ‘x’.

Here in this problem, we are given an equation of inverse trigonometric function, i.e. $2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\operatorname{co} {\text{sec }}x} \right)$ . And we need to solve this equation and find the value of the variable $x$ .
Before starting with the solution to this problem, we should understand the concepts of the inverse trigonometric function. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. They are also termed as arcus functions, antitrigonometric functions, or cyclometric functions. These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios.
So we can represent the given equation as:
Let us assume that: $2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\operatorname{co} {\text{sec }}x} \right) = m$
Now we can represent the expressions on RHS and LHS as:
$\Rightarrow m = 2{\tan ^{ - 1}}\left( {\cos x} \right){\text{ and }}m = {\tan ^{ - 1}}\left( {2\operatorname{co} {\text{sec }}x} \right)$
Therefore,
$\Rightarrow \dfrac{m}{2} = {\tan ^{ - 1}}\left( {\cos x} \right){\text{ and }}m = {\tan ^{ - 1}}\left( {2\operatorname{co} {\text{sec }}x} \right)$
From the definition of the inverse tangent function, we the tangent of LHS will be equal to the tangent of the tangent of the expression on RHS
$\Rightarrow \tan \left( {\dfrac{m}{2}} \right) = \tan \left( {{{\tan }^{ - 1}}\left( {\cos x} \right)} \right){\text{ = }}\cos x$
Also, we have:
$\Rightarrow \tan \left( m \right) = \tan \left( {{{\tan }^{ - 1}}\left( {2{\text{cosec }}x} \right)} \right) = 2{\text{cosec }}x$
Therefore, we get the value of the tangent of ‘m’ and tangent of half of ‘m’. Now we can easily use the tangent half-angle formula, i.e.
$\Rightarrow \tan m = \dfrac{{2\tan \dfrac{m}{2}}}{{1 - {{\tan }^2}\dfrac{m}{2}}}$
Let’s substitute the above found values in the above formula:
$\Rightarrow \tan m = \dfrac{{2\tan \dfrac{m}{2}}}{{1 - {{\tan }^2}\dfrac{m}{2}}} \Rightarrow 2\cos ec{\text{ }}x = \dfrac{{2 \times \cos x}}{{1 - {{\left( {\cos x} \right)}^2}}}$
As we know that the sum of squared of sine and cosine is one, i.e. ${\sin ^2}x + {\cos ^2}x = 1$
We can use this property to change the denominator of the above expression, i.e. we can substitute $1 - {\cos ^2}x = {\sin ^2}x$
Therefore, this will give us:
$\Rightarrow 2\cos ec{\text{ }}x = \dfrac{{2 \times \cos x}}{{1 - {{\left( {\cos x} \right)}^2}}} = \dfrac{{2\cos x}}{{{{\sin }^2}x}}$
Now changing the $\cos ec{\text{ }}x$ into $\dfrac{1}{{\sin x}}$ and dividing both sides by $2$ , we get:
$\Rightarrow \dfrac{{\dfrac{2}{{\sin x}}}}{2} = \dfrac{{\dfrac{{2\cos x}}{{{{\sin }^2}x}}}}{2} \Rightarrow \dfrac{1}{{\sin x}} = \dfrac{{\cos x}}{{{{\sin }^2}x}}$
Transposing RHS to LHS, we get:
$\Rightarrow \dfrac{{{{\sin }^2}x}}{{\sin x \times \cos }} = 1 \Rightarrow \dfrac{{\sin x}}{{\cos x}} = 1$
Since we know that tangent function is the ratio of sine and cosine function, i.e. $\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\Rightarrow \dfrac{{\sin x}}{{\cos x}} = \tan x = 1$
Therefore, we get $\tan x = 1$ that means that $\sin x = \cos x$ , which is only possible when $x = {45^ \circ }{\text{ or }}\dfrac{\pi }{4}$
Hence, the option (A) is the correct answer.

Note: In questions like this, we should use the trigonometric properties to transform the only given expression to find the value. An alternative approach can be taken by obtaining an equation $\sin x = \cos x$ and then check the given option, i.e. $x = \dfrac{\pi }{4},\dfrac{\pi }{2},\dfrac{\pi }{6}$ which satisfies the above equation. This will eliminate the step of using the tangent function.