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

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(A) આપેલ સમીકરણ: $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$.પદોને ગોઠવતા: $\operatorname{c

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) આપેલ સમીકરણ: $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$.પદોને ગોઠવતા: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)$.નિત્યસમ $\cos^2	heta+\sin^2	heta=1$ નો ઉપયોગ કરતા: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=1$.આ સમીકરણ ત્યારે જ શક્ય છે જ્યારે $\operatorname{cosec}^2(\alpha+\beta)=1$ અને $\sin^2(2\alpha-\beta)=0$ હોય.તેથી,$\alpha+\beta = \frac{\pi}{2}$ અને $2\alpha-\beta = 0$.સમીકરણો ઉકેલતા: $\alpha = \frac{\pi}{6}$ અને $\beta = \frac{\pi}{3}$.આમ,$\sin(\alpha-\beta) = \sin(\frac{\pi}{6}-\frac{\pi}{3}) = \sin(-\frac{\pi}{6}) = -\frac{1}{2}$.

જો $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$ જ્યાં $\alpha, \beta \in (0, \frac{\pi}{2})$ હોય,તો $\sin(\alpha-\beta)$ ની કિંમત શોધો.

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