Question

The value of ${\tan }^{-1}\left(\frac{\sin 1-1}{\cos 1}\right)$ equal -

• A. $0$
• B. $1-\frac{\pi }{2}$
• C. $\frac{\pi }{2}-1$
• D. $\frac{1}{2}-\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{1}{2}-\frac{\pi }{4}$

$\frac{\sin 1-1}{\cos 1}=\frac{2\sin \frac{1}{2}\cos \frac{1}{2}-{\sin }^2\frac{1}{2}-{\cos }^2\frac{1}{2}}{{\cos }^2\frac{1}{2}-{\sin }^2\frac{1}{2}}$

$=\frac{(\sin \frac{1}{2}-\cos \frac{1}{2}{)}^2}{(\sin \frac{1}{2}-\cos \frac{1}{2})(\sin \frac{1}{2}+\cos \frac{1}{2})}$

$=\frac{\sin \frac{1}{2}-\cos \frac{1}{2}}{\sin \frac{1}{2}+\cos \frac{1}{2}}$

$=\frac{\tan \frac{1}{2}-1}{1+\tan \frac{1}{2}}$

$=\frac{\tan \frac{1}{2}-\tan \frac{\pi }{4}}{1+\tan \frac{\pi }{4}\tan \frac{1}{2}}$

$=\tan (\frac{1}{2}-\frac{\pi }{4})$

$⟹{\text{tan}}^{-1}(\frac{\sin 1-1}{\cos 1})=\frac{1}{2}-\frac{\pi }{4}$