Question

If in a right angled $\Delta ABC$,4 sin A cos B – 1 = 0 and tan A is real, then

• A. angles are in GP.
• B. angles are in HP.
• C. angles are in AP.
• D. None of these

Answer 1

The correct option is C angles are in AP.

Since, 4 sin A cos B – 1 = 0. So, A and B cannot be $\frac{\pi }{2}${As, if $B=\frac{\pi }{2}$, then cos B =0 and if $A=\frac{\pi }{2}$, tan A is defined}
So,$C=\frac{\pi }{2}\Rightarrow B=\frac{\pi }{2}-A\Rightarrow 4sinAcos(\frac{\pi }{2}-A)=1$
$\Rightarrow si{n}^{2}A=\frac{1}{4}\Rightarrow sinA=\frac{1}{2}\Rightarrow A=\frac{\pi }{6}\Rightarrow B=\frac{\pi }{3}$
So, the angle are $\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2}$ which are in AP.