If operatorname{cosec}^2(α+β)-sin ^2(β-α)+sin ^2(2 α-β)=cos ^2(α-β) where α, β ∈left(0,

English

Gujarati

Hindi

Trigonometrical Equations

normal

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

$-\frac{1}{2}$

$\frac{1}{2}$

$\frac{-\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}$

(KVPY-2009)

(a)

$\operatorname{cosec}^2(\alpha+\beta)-\sin ^2(\beta-\alpha)$

$+\sin ^2(2 \alpha-\beta)=\cos ^2(\alpha-\beta)$

$\Rightarrow \operatorname{cosec}^2(\alpha+\beta)+\sin ^2(2 \alpha-\beta)$

$=\cos ^2(\alpha-\beta)+\sin ^2(\alpha-\beta)$

$\Rightarrow \operatorname{cosec}^2(\alpha+\beta)+\sin ^2(2 \alpha-\beta)=1$

lt is possible only $\operatorname{cosec}^2(\alpha+\beta)=1$ and $\sin ^2(2 \alpha-\beta)=0$

$\therefore \quad \alpha+\beta=\frac{\pi}{2}$ and $2 \alpha-\beta=0$

On solving these equations, we get

$\alpha=\frac{\pi}{6} \text { and } \beta=\frac{\pi}{3}$

$\therefore \sin (\alpha-\beta)=\sin \left(\frac{\pi}{6}-\frac{\pi}{3}\right)=\sin \left(-\frac{\pi}{6}\right)=-\frac{1}{2}$

Standard 11

Mathematics

normal

hard

(JEE MAIN-2021)

hard

normal

(KVPY-2009)

medium