Question

Value of ${\tan }^{-1}\left[\frac{\sin \left(2\right)-1}{\cos \left(2\right)}\right]$ is

• A. $\left(\frac{\mathrm{\pi }}{2}\right)-1$
• B. $1-\left(\frac{\mathrm{\pi }}{4}\right)$
• C. $2-\left(\frac{\mathrm{\pi }}{2}\right)$
• D. $\left(\frac{\mathrm{\pi }}{4}\right)-1$

Answer 1

The correct option is B

$1-\left(\frac{\mathrm{\pi }}{4}\right)$

Explanation for the correct option

Step 1: Solve for the required value

Given: ${\tan }^{-1}\left[\frac{\sin \left(2\right)-1}{\cos \left(2\right)}\right]$

As we know that $\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right),\cos \left(2x\right)={\cos }^2\left(x\right)-{\sin }^2\left(x\right),{\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1$

Putting $x=1$

$$\begin{gathered} ={\tan }^{-1}\left[\frac{2\sin \left(1\right)\cos \left(1\right)-\left[\left({\sin }^2\right(1)+{\cos }^2(1)\right]}{{\cos }^2\left(1\right)-{\sin }^2\left(1\right)}\right] \\ ={\tan }^{-1}\left[\frac{\left({\sin }^2\right(1)+{\cos }^2(1\left)\right)-2\sin \left(1\right)\cos \left(1\right)}{{\sin }^2\left(1\right)-{\cos }^2\left(1\right)}\right] \\ ={\tan }^{-1}\left[\frac{{\left[\sin \left(1\right)-\cos \left(1\right)\right]}^2}{\left[\sin \left(1\right)-\cos \left(1\right)\right]\left[\sin \left(1\right)+\cos \left(1\right)\right]}\right]\left[\because a^2-b^2=(a-b)(a+b),a^2+b^2-2ab={(a-b)}^2\right] \\ ={\tan }^{-1}\left[\frac{\left[\sin \left(1\right)-\cos \left(1\right)\right]}{\left[\sin \left(1\right)+\cos \left(1\right)\right]}\right] \end{gathered}$$

Step 2: Solve further for the required value

Dividing numerator and denominator by $\cos \left(1\right)$

$$\begin{gathered} ={\tan }^{-1}\left[\frac{\left({\displaystyle \frac{\sin \left(1\right)}{\cos \left(1\right)}}\right)-\left({\displaystyle \frac{\cos \left(1\right)}{\cos \left(1\right)}}\right)}{{\displaystyle \frac{\sin \left(1\right)}{\cos \left(1\right)}}+{\displaystyle \frac{\cos \left(1\right)}{\cos \left(1\right)}}}\right] \\ ={\tan }^{-1}\left[\frac{\left({\displaystyle \tan \left(1\right)}\right)-\left({\displaystyle 1}\right)}{{\displaystyle \tan \left(1\right)+1}}\right] \\ ={\tan }^{-1}\left[\frac{\tan \left(1\right)-\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{{\displaystyle 1+\tan \left(1\right)\tan \left(\frac{\mathrm{\pi }}{4}\right)}}\right]\left[\because \tan \left(\frac{\mathrm{\pi }}{4}\right)=1\right] \end{gathered}$$

As we know that $\frac{\tan \left(x\right)-\tan \left(y\right)}{1+\tan \left(x\right)\tan \left(y\right)}=\tan (x-y)$

Putting $x=1,y=\frac{\mathrm{\pi }}{4}$

$$\begin{gathered} ={\tan }^{-1}\left(\tan \left(1-\frac{\mathrm{\pi }}{4}\right)\right) \\ =1-\frac{\mathrm{\pi }}{4} \end{gathered}$$

Hence, option(B) i.e. $1-\left(\frac{\mathrm{\pi }}{4}\right)$ is correct.