# If we are given $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right)$ , then find the value of x.(a) $\dfrac{\pi }{2}$ .(b) $\pi$ .(c) $\dfrac{\pi }{6}$ .(d) $\dfrac{\pi }{3}$ .

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Hint: We convert $2{{\tan }^{-1}}\left( \cos x \right)$ to ${{\tan }^{-1}}\left( \cos x \right)+{{\tan }^{-1}}\left( \cos x \right)$ for applying ${{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ . After applying the formula, we obtain a value for $\cos x$ . Using the value of $\cos x$ , we find the value of x. We verify by substituting the value of ‘x’ in the original given equation.

We are given $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right)$ , and we need to find the value of ‘x’.
We have ${{\tan }^{-1}}\left( \cos x \right)+{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right)$ .
We know that ${{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ for ab<1.
We know the value of cos x lies in the interval $\left[ -1,1 \right]$ and the value of $\cos x\times \cos x$ lies in the interval $\left[ 0,1 \right]$ .
So, we have ${{\tan }^{-1}}\left( \dfrac{\cos x+\cos x}{1-\left( \cos x\times \cos x \right)} \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right)$ .
We have ${{\tan }^{-1}}\left( \dfrac{2\cos x}{1-{{\cos }^{2}}x} \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right)$ .
We know from the trigonometric identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ .
We have ${{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right)$ .
We know that $\operatorname{cosec}x=\dfrac{1}{\sin x}$ .
We have ${{\tan }^{-1}}\left( 2\cos x.{{\operatorname{cosec}}^{2}}x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right)$ .
We know that if ${{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( y \right)$ , then x = y.
We have $2\cos x.{{\operatorname{cosec}}^{2}}x={{\operatorname{cosec}}^{2}}x$ .
We have 2cos x = 1.
We have $\cos x=\dfrac{1}{2}$ .
We have $x={{\cos }^{-1}}\left( \dfrac{1}{2} \right)$.
We have $x={{60}^{o}}$ .
We know that $\pi ={{180}^{o}}$ .
We have $x={{60}^{o}}\times \dfrac{{{180}^{o}}}{{{180}^{o}}}$ .
We have $x=\dfrac{\pi }{3}$.
We got the value of ‘x’ as $\dfrac{\pi }{3}$ .
Now, we verify the obtained result.
We have $2{{\tan }^{-1}}\left( \cos \left( \dfrac{\pi }{3} \right) \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}\left( \dfrac{\pi }{3} \right) \right)$ .
We know that $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ and $\operatorname{cosec}\left( \dfrac{\pi }{3} \right)=\dfrac{2}{\sqrt{3}}$ .
So, we have $2{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( {{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}} \right)$ .
We have ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right)$ .
Since, $\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}$ , which is less than 1, we use ${{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ .
We have ${{\tan }^{-1}}\left( \dfrac{\dfrac{1}{2}+\dfrac{1}{2}}{1-\left( \dfrac{1}{2}\times \dfrac{1}{2} \right)} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right)$ .
We have ${{\tan }^{-1}}\left( \dfrac{1}{1-\left( \dfrac{1}{4} \right)} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right)$ .
We have ${{\tan }^{-1}}\left( \dfrac{1}{\dfrac{3}{4}} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right)$ .
We have ${{\tan }^{-1}}\left( \dfrac{4}{3} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right)$ .
∴ We have proved that the value of ‘x’ is $\dfrac{\pi }{3}$ .
So, the correct answer is “Option D”.

Note: We should not every time use ${{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ as it depends on the value of ab. We always verify the value obtained if the given equation of inverse trigonometric functions consists of arctanx $\left( {{\tan }^{-1}}x \right)$ . Sometimes, the obtained value may not satisfy the given equation in such cases we declare that ‘x’ doesn’t have a solution.