tan 81^{circ}-tan 63^{circ}-tan 27^{circ}+tan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We have the expression: $	an 81^{\circ} + 	an 9^{\circ} - (	an 63^{\circ} + 	an 27^{\circ})$Using the identity $	an A + 	an B = \frac{\sin(A

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) We have the expression: $	an 81^{\circ} + 	an 9^{\circ} - (	an 63^{\circ} + 	an 27^{\circ})$Using the identity $	an A + 	an B = \frac{\sin(A+B)}{\cos A \cos B}$:$	an 81^{\circ} + 	an 9^{\circ} = \frac{\sin(81^{\circ}+9^{\circ})}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{\sin 90^{\circ}}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} = \frac{2}{\sin 18^{\circ}}$Similarly,$	an 63^{\circ} + 	an 27^{\circ} = \frac{\sin(63^{\circ}+27^{\circ})}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{\sin 90^{\circ}}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} = \frac{2}{\sin 54^{\circ}}$Now,the expression becomes $\frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}} = 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}} ight)$Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$\sin 54^{\circ} - \sin 18^{\circ} = 2 \cos 36^{\circ} \sin 18^{\circ}$So,$2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}} ight) = 2 	imes 2 = 4$

$\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$ equals to

Similar Questions

The number of real roots of the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$ is:

If $\frac{\tan (\alpha+\beta-\gamma)}{\tan (\alpha-\beta+\gamma)}=\frac{\tan \gamma}{\tan \beta}$ and $\beta \neq \gamma$,then the value of $\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma$ is

If $\sin A + \sin B = \frac{1}{2}$ and $\cos A + \cos B = 1$,then $\sin \left(\frac{A-B}{2}\right)$ equals

If $\frac{\cos x}{\cos (x-2 y)}=\lambda$,then $\tan (x-y) \tan y$ is equal to

If $\tan(\pi \sin \theta) = \cot(\pi \cos \theta)$,then $\left| \cot \left( \theta - \frac{\pi}{4} \right) \right|$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module