यदि alpha = frac{sin^3 x}{cos^2 x},beta = fra | vedclass

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

Question

(B) दिया गया है,$\alpha = \frac{\sin^3 x}{\cos^2 x}$ और $\beta = \frac{\cos^3 x}{\sin^2 x}$.$\alpha \sin x + \beta \cos x + 3 = \frac{\sin^4 x}{\cos^2

Vedclass · Accepted Answer

$\frac{4}{(k^2-1)^2}$

Vedclass · Answer

(B) दिया गया है,$\alpha = \frac{\sin^3 x}{\cos^2 x}$ और $\beta = \frac{\cos^3 x}{\sin^2 x}$.$\alpha \sin x + \beta \cos x + 3 = \frac{\sin^4 x}{\cos^2 x} + \frac{\cos^4 x}{\sin^2 x} + 3$.$= \frac{\sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.सर्वसमिका $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ का उपयोग करते हुए,जहाँ $a = \sin^2 x$ और $b = \cos^2 x$:$= \frac{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.$= \frac{(\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$.दिया गया है $\sin x + \cos x = k$,दोनों पक्षों का वर्ग करने पर $1 + 2 \sin x \cos x = k^2$,इसलिए $\sin x \cos x = \frac{k^2 - 1}{2}$.अतः,$\frac{1}{\sin^2 x \cos^2 x} = \frac{1}{(\frac{k^2 - 1}{2})^2} = \frac{4}{(k^2 - 1)^2}$.

यदि $\alpha = \frac{\sin^3 x}{\cos^2 x}$,$\beta = \frac{\cos^3 x}{\sin^2 x}$ और $\sin x + \cos x = k$ है,तो $\alpha \sin x + \beta \cos x + 3 = $

Similar Questions

यदि $\sin \theta + \sin^2 \theta = 1$ और $\cos^{12} \theta + a \cos^{10} \theta + b \cos^8 \theta + c \cos^6 \theta + d = 0$ है,तो:

व्यंजक $\frac{2(\sin 1^{\circ} + \sin 2^{\circ} + \sin 3^{\circ} + \dots + \sin 89^{\circ})}{2(\cos 1^{\circ} + \cos 2^{\circ} + \dots + \cos 44^{\circ}) + 1}$ का मान क्या है?

यदि $\operatorname{Sinh}^{-1} x = \operatorname{Cosh}^{-1} y = \log(1+\sqrt{2})$ है, तो $\operatorname{Tan}^{-1}(x+y) = $

यदि $\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ} = m$ और $\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ} = n$ है,तो $\frac{n}{m}$ का मान ज्ञात कीजिए।

व्यंजक $\frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)} - \frac{1}{4} \sin 2 \alpha \left[\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right]$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module