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

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

Question

(D) हमारे पास है,$	an 81^{\circ} - 	an 63^{\circ} - 	an 27^{\circ} + 	an 9^{\circ}$$= (	an 81^{\circ} + 	an 9^{\circ}) - (	an 63^{\circ} + 	an

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) हमारे पास है,$	an 81^{\circ} - 	an 63^{\circ} - 	an 27^{\circ} + 	an 9^{\circ}$$= (	an 81^{\circ} + 	an 9^{\circ}) - (	an 63^{\circ} + 	an 27^{\circ})$$= (\cot 9^{\circ} + 	an 9^{\circ}) - (\cot 27^{\circ} + 	an 27^{\circ})$$= \left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}} + \frac{\sin 9^{\circ}}{\cos 9^{\circ}}ight) - \left(\frac{\cos 27^{\circ}}{\sin 27^{\circ}} + \frac{\sin 27^{\circ}}{\cos 27^{\circ}}ight)$$= \left(\frac{\cos^2 9^{\circ} + \sin^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}ight) - \left(\frac{\cos^2 27^{\circ} + \sin^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}ight)$$= \left(\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}ight) - \left(\frac{2}{2 \sin 27^{\circ} \cos 27^{\circ}}ight)$$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$$= 2 \left(\frac{1}{\sin 18^{\circ}} - \frac{1}{\cos 36^{\circ}}ight)$$= 2 \left(\frac{4}{\sqrt{5} - 1} - \frac{4}{\sqrt{5} + 1}ight)$$= 8 \left(\frac{\sqrt{5} + 1 - (\sqrt{5} - 1)}{5 - 1}ight)$$= 8 \left(\frac{2}{4}ight) = 4$

$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$ का मान है

Similar Questions

फलन $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$ का आलेख है

मान लीजिए $A$ और $B$ कथन हैं:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
यदि $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$ है,तो:

यदि $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ है,तो $\cos ^2 48^{\circ} - \sin ^2 12^{\circ}$ का मान क्या है?

$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ का मान ज्ञात कीजिए।

यदि $\sum_{r=1}^{13} \left\{ \frac{1}{\sin \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{r\pi}{6}\right)} \right\} = a\sqrt{3} + b$,जहाँ $a, b \in \mathbb{Z}$,तो $a^2 + b^2$ का मान ज्ञात कीजिए:

Vedclass Test Series

Exam Paper Generator

Online Exam Module