tan alpha + 2 tan 2 alpha + 4 tan 4 alpha + 8 | vedclass

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

Question

(C) हम सर्वसमिका $	an 	heta - \cot 	heta = -2 \cot 2 	heta$ का उपयोग करते हैं। दी गई व्यंजक: $S = 	an \alpha + 2 	an 2 \alpha + 4 	an 4 \alpha

Vedclass · Accepted Answer

$\cot \alpha$

Vedclass · Answer

(C) हम सर्वसमिका $	an 	heta - \cot 	heta = -2 \cot 2 	heta$ का उपयोग करते हैं। दी गई व्यंजक: $S = 	an \alpha + 2 	an 2 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$. $	an \alpha = \cot \alpha - 2 \cot 2 \alpha$ प्रतिस्थापित करने पर: $S = (\cot \alpha - 2 \cot 2 \alpha) + 2 	an 2 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha + 2(	an 2 \alpha - \cot 2 \alpha) + 4 	an 4 \alpha + 8 \cot 8 \alpha$ चूंकि $	an 2 \alpha - \cot 2 \alpha = -2 \cot 4 \alpha$: $S = \cot \alpha + 2(-2 \cot 4 \alpha) + 4 	an 4 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha - 4 \cot 4 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha + 4(	an 4 \alpha - \cot 4 \alpha) + 8 \cot 8 \alpha$ चूंकि $	an 4 \alpha - \cot 4 \alpha = -2 \cot 8 \alpha$: $S = \cot \alpha + 4(-2 \cot 8 \alpha) + 8 \cot 8 \alpha$ $S = \cot \alpha - 8 \cot 8 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha$.

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

Similar Questions

$(4 \cos^2 9^{\circ} - 3)(4 \cos^2 27^{\circ} - 3) = $

यदि $\frac{5\pi}{2} < x < 3\pi$ है,तो व्यंजक $\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}$ का मान क्या है?

यदि $\tan A < 0$ और $\tan 2A = -\frac{4}{3}$ है,तो $\cos 6A =$

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

$\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module