If cos (alpha+beta)=0, then sin (alpha-beta) | vedclass

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Question

(A) Given,$\cos (\alpha+\beta)=0=\cos 90^{\circ}$ $\left[\because \cos 90^{\circ}=0\right]$$\Rightarrow \alpha+\beta=90^{\circ}$$\Rightarrow \alpha=90

Vedclass · Accepted Answer

$\cos 2 \beta$

Vedclass · Answer

(A) Given,$\cos (\alpha+\beta)=0=\cos 90^{\circ}$ $\left[\because \cos 90^{\circ}=0ight]$$\Rightarrow \alpha+\beta=90^{\circ}$$\Rightarrow \alpha=90^{\circ}-\beta$ ......$(i)$Now,$\sin (\alpha-\beta)=\sin (90^{\circ}-\beta-\beta)$ [substituting the value from Eq. $(i)$]$= \sin (90^{\circ}-2 \beta)$$= \cos 2 \beta$ $\left[\because \sin (90^{\circ}-	heta)=\cos 	hetaight]$Hence,$\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.

Part $I$	Part $II$
$1.$ $\cos(90^\circ - \theta)$	$a.$ $\sec \theta$
$2.$ $\cot(90^\circ - \theta)$	$b.$ $\sin \theta$
$3.$ $\operatorname{cosec}(90^\circ - \theta)$	$c.$ $1$
	$d.$ $\tan \theta$

If $\cos (\alpha+\beta)=0,$ then $\sin (\alpha-\beta)$ can be reduced to

Similar Questions

Which of the following groups truly matches the data of Part $I$ with the data of Part $II$?

Part $I$ Part $II$

$1.$ $\cos(90^\circ - \theta)$ $a.$ $\sec \theta$

$2.$ $\cot(90^\circ - \theta)$ $b.$ $\sin \theta$

$3.$ $\operatorname{cosec}(90^\circ - \theta)$ $c.$ $1$

$d.$ $\tan \theta$

If $\cos \theta = \frac{1}{\sqrt{2}},$ then $\theta = \ldots$ (in $^\circ$)

Simplify $(1+\tan ^{2} \theta)(1-\sin \theta)(1+\sin \theta)$

$\frac{\sin 60^{\circ} + \cos 30^{\circ}}{1 + \sin 30^{\circ} + \cos 60^{\circ}} = \dots$

If $\tan \theta = \frac{5}{12}$,then $\cos \theta = \ldots$

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