Question

If $\cos (\alpha -\beta )=\sin (\alpha +\beta )$, then $\frac{1}{1-\sin \alpha }=$

• A. $\frac{\sqrt{2}}{\sqrt{2}-1}$
• B. $\frac{2}{\sqrt{2}-1}$
• C. $1$
• D. None of these

Answer 1

The correct option is A $\frac{\sqrt{2}}{\sqrt{2}-1}$

$If\cos (\alpha -\beta )=\sin (\alpha +\beta ).Then\frac{1}{1-\sin \alpha }\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \cos \alpha \cos \beta +\sin \alpha \sin \beta -\sin \alpha \cos \beta -\cos \alpha \sin \beta =0\cos \alpha \left(\cos \beta -\sin \beta \right)-\sin \alpha \left(\cos \beta -\sin \beta \right)=0\left(\cos \alpha -\sin \alpha \right)\left(\cos \beta -\sin \beta \right)=0\cos \alpha -\sin \alpha =0\cos \alpha =\sin \alpha \cos \alpha =\cos (90-\alpha )\alpha =90-\alpha 2\alpha =90\alpha =45°Then,\frac{1}{1-\sin {45}^o}=\frac{1}{1-\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{2}-1}$