Question

Question 5
If $cos(\alpha +\beta )=0$, then $sin(\alpha -\beta )$ can be reduced to

(A) $cos\beta$
(B) $cos2\beta$
(C) $sin\alpha$
(D) $sin2\alpha$

Answer 1

The answer is (B).

Given, $cos(\alpha +\beta )=0=cos{90}^{\circ }\lfloor \because cos{90}^{\circ }=0\rfloor$

$\Rightarrow \alpha +\beta ={90}^{\circ }$

$\Rightarrow \alpha ={90}^{\circ }-\beta$ .........(i)

Now, $sin(\alpha -\beta )=sin({90}^{\circ }-\beta -\beta )$ [substituting the value from Eq. (i)]

$=sin({90}^{\circ }-2\beta )$

$=cos2\beta \lfloor \because sin({90}^{\circ }-\theta )\rfloor$

$\therefore$ $sin(\alpha -\beta )$ can be reduced to $cos2\beta$.