Browse Questions

If $\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$ then the value of $\cot\alpha\tan\beta$ is

$(a)\;-1\qquad(b)\;0\qquad(c)\;1\qquad(d)\;None\;of\;these$

Can you answer this question?

Given :
$\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$
$\Rightarrow \cos(\alpha+\beta)=-1$
$\alpha+\beta=-1$
$\alpha+\beta=(2n+1)\pi$
$1+\cot \alpha\tan\beta=1+\large\frac{\cos\alpha}{\sin\alpha}.\large\frac{\sin\beta}{\cos\beta}$
$\Rightarrow \large\frac{\sin\alpha\sin\beta+\cos\alpha\sin\beta}{\sin\alpha\cos\beta}$
$\Rightarrow \large\frac{\sin(\alpha+\beta)}{\sin\alpha\cos\beta}$
$\Rightarrow \large\frac{\sin((2n+1)\pi)}{\sin\alpha\cos\beta}$
$\Rightarrow 0$
$1+\cot \alpha+\tan\beta=0$
$\cot \alpha+\tan\beta=-1$
Hence (a) is the correct answer.
answered Oct 15, 2013