If operatorname{cosec}^2(alpha+beta)-sin^2(be | vedclass

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

Question

(A) Given the equation: $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$.Rearranging the terms,w

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) Given the equation: $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$.Rearranging the terms,we get: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)+\sin^2(\beta-\alpha)$.Since $\sin^2(\beta-\alpha) = \sin^2(\alpha-\beta)$,the equation becomes: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)$.Using the identity $\cos^2	heta+\sin^2	heta=1$,we have: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=1$.Since $\operatorname{cosec}^2(\alpha+\beta) \geq 1$ and $\sin^2(2\alpha-\beta) \geq 0$,the sum can be $1$ only if $\operatorname{cosec}^2(\alpha+\beta)=1$ and $\sin^2(2\alpha-\beta)=0$.This implies $\alpha+\beta = \frac{\pi}{2}$ and $2\alpha-\beta = 0$.Adding the two equations: $3\alpha = \frac{\pi}{2} \implies \alpha = \frac{\pi}{6}$.Substituting $\alpha$ back: $\beta = 2\alpha = \frac{\pi}{3}$.Thus,$\sin(\alpha-\beta) = \sin(\frac{\pi}{6}-\frac{\pi}{3}) = \sin(-\frac{\pi}{6}) = -\frac{1}{2}$.

If $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\sin(\alpha-\beta)$ is equal to

Similar Questions

$\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$ is equal to

If $\sin A + \sin 2A = x$ and $\cos A + \cos 2A = y,$ then $({x^2} + {y^2})({x^2} + {y^2} - 3) = $

If $\alpha, \beta$ are acute angles such that $\frac{\sin \alpha}{\sin \beta} = \frac{6}{5}$ and $\frac{\cos \alpha}{\cos \beta} = \frac{9}{5 \sqrt{5}}$,then $\sin \alpha = $

If $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$,then $f_1(x) + f_2(x) - f_3(x) =$

One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module