Question

In $△ABC$, if $C={90}^o,1+\sin A-\sin B=$

• A. $\frac{r}{2R}$
• B. $\frac{2r}{R}$
• C. $\frac{2r_1}{R}$
• D. $\frac{r_1}{R}$

Answer 1

Answer: (A)

$\frac{r}{2R}$

Since $\angle A+\angle B+\angle C={180}^{\circ }$

And $\angle C={90}^{\circ }$

$\angle A+\angle B={90}^{\circ }$

$\cos (A+B)=\cos {90}^{\circ }=0$

$\cos (A+B)=\cos A\cos B-\sin A\sin B=0$

So, $\tan A\tan B=1$

Maximum value of $\tan A=\tan B=1$

When $A=B=\pi /4$

Thus $\sin A\sin B=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{1}{2}$