Question

Assertion : lf $x=\sin (\alpha -\beta )\sin (\gamma -\delta )$ ,$y=\sin (\beta -\gamma )\sin (\alpha -\delta )$, $z=\sin (\gamma -\alpha )\sin (\beta -\delta )$, then $x+y+z=0$
Reason : $2\sin A\sin B=\cos (A-B)+\cos (A+B)$

• A. A is true, R is true and R is correct explanation of A
• B. A is true, R is true and R is not correct explanation of A
• C. A is true, R is false
• D. A is false, R is true.

Answer 1

The correct option is C A is true, R is false

Assertion is true as

$x=\sin (\alpha -\beta )\sin (\gamma -\beta )=(\sin \alpha \cos \beta -\cos \alpha \sin beta)(\sin \gamma \cos \delta -\cos \gamma sin\delta )$

$=\sin \alpha \cos \beta sin\gamma \cos \delta -\sin \alpha \cos \beta \cos \gamma \sin \delta -\cos \alpha \sin \beta \sin \gamma \cos \gamma +\cos \alpha \sin \beta \cos \gamma \sin \delta$

$y=\sin (\beta -\gamma )\sin (\alpha -\delta )=(\sin \beta \cos \gamma -\cos \beta \sin \gamma )(\sin \alpha \cos \delta -\cos \alpha sin\gamma )$

$=sin\alpha \sin \beta \cos \gamma \cos \delta -\cos \alpha \sin \beta \cos \gamma \sin \delta -\sin \alpha \cos \beta \sin \gamma \cos \delta +cos\alpha \cos \beta \sin \gamma \sin \delta$

$z=\sin (\gamma -\alpha )\sin (\beta -\gamma )=(\sin \gamma \cos \alpha -\cos \gamma \sin \alpha )(\sin \beta \cos \delta -\cos \beta \sin \delta )$

$=\cos \alpha \sin \beta \sin \gamma \cos \delta -\cos \alpha \cos \beta \sin \gamma \sin \delta -\sin \alpha \sin \beta \cos \gamma \cos \delta +\sin \alpha \cos \beta \cos \gamma \sin \delta$

$x+y+z=0$

Reason is false as

$2\sin A\sin B=\cos (A-B)-\cos (A+B)$