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The value of ${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)$ is equal to
(a) $\dfrac{\pi }{2}-1$
(b) $2-\dfrac{\pi }{2}$
(c) $1-\dfrac{\pi }{4}$
(d) $\dfrac{\pi }{4}-1$

Last updated date: 18th Jun 2024
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Hint: In order to solve this problem, we need to know multiple numbers of trigonometric identities and formula. The formulas we need to know are as follows, $\sin 2x=2\sin x\times \cos x$, $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$, ${{\sin }^{2}}1+{{\cos }^{2}}1=1$ , ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a.\tan b}$ and $\tan \left( -x \right)=-\tan x$.

Complete step-by-step solution:
We are given the expression and we need to find the value of it.
The expression is ${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)$ .
In order to simplify this, we need to know certain identities.
The identities are $\sin 2x=2\sin x\times \cos x$
And $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$
Using these identities, we get,
${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{2\sin 1.\cos 1-1}{{{\cos }^{2}}1-{{\sin }^{2}}1} \right)$
We can now use the rule that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the denominator, we get,
${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{2\sin 1.\cos 1-1}{\left( \cos 1+\sin 1 \right)\left( \cos 1-\sin 1 \right)} \right)$
We can write 1 as ${{\sin }^{2}}1+{{\cos }^{2}}1$ because of the identity that ${{\sin }^{2}}1+{{\cos }^{2}}1=1$ .
The equation becomes,
${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{2\sin 1.\cos 1-\left( {{\sin }^{2}}1+{{\cos }^{2}}1 \right)}{\left( \cos 1+\sin 1 \right)\left( \cos 1-\sin 1 \right)} \right)$
In the numerator, we can combine all the terms using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ ,
We get,
  ${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-{{\left( \cos 1-\sin 1 \right)}^{2}}}{\left( \cos 1+\sin 1 \right)\left( \cos 1-\sin 1 \right)} \right)$
We can cancel few of the terms form numerator as well as the denominator.
After solving we get,
\[{{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-\left( \cos 1-\sin 1 \right)}{\left( \cos 1+\sin 1 \right)} \right)\]
Taking the cos 1 common from the numerator we get,
  & {{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-\cos 1\left( 1-\tan 1 \right)}{\cos 1\left( 1+\tan 1 \right)} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{-\left( 1-\tan 1 \right)}{\left( 1+\tan 1 \right)} \right)
We knew that $\tan \dfrac{\pi }{4}=1$ , substituting we get,
\[{{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-\left( \tan \dfrac{\pi }{4}-\tan 1 \right)}{\left( 1+\tan \dfrac{\pi }{4}.\tan 1 \right)} \right)\]
We can see that \[\dfrac{\left( \tan \dfrac{\pi }{4}-\tan 1 \right)}{\left( 1+\tan \dfrac{\pi }{4}.\tan 1 \right)}={{\tan }^{-1}}\left( \dfrac{\pi }{4}-1 \right)\].
We can use identity $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a.\tan b}$ .
Substituting we get,
\[{{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( -\tan \left( \dfrac{\pi }{4}-1 \right) \right)\]
We need to take the negative inside.
We must know the property $\tan \left( -x \right)=-\tan x$ .
Solving this we get,
  & {{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \tan \left( 1-\dfrac{\pi }{4} \right) \right) \\
 & =1-\dfrac{\pi }{4}
Hence, the correct option is (c).

Note: In this problem, the main trick is to know which formula to apply when. Also, many tend to make the wrong approach. Another approach that is usually taken is as follows,\[\begin{align}
  & {{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{\sin 2}{\cos 2}-\dfrac{1}{\cos 2} \right) \\
 & ={{\tan }^{-1}}\left( \tan 2-\sec 2 \right) \\
 & =2-{{\tan }^{-1}}\left( \sec 2 \right)
But this is not the correct approach to take.