Question

If $sinA=\frac{3}{5},\tan B=\frac{1}{2}$ and $\frac{\pi }{2}<A<\pi <B<\frac{3\pi }{2}$, then the value of $8\tan A-\sqrt{5}secB=$

• A. $5/2$
• B. $-5/2$
• C. $-7/2$
• D. $7/2$

Answer 1

The correct option is C $-7/2$

Given $\frac{\pi }{2}<A<\pi ,⟹\tan A<0$

$\sin A=\frac{3}{5}⟹\tan A=-\frac{3}{\sqrt{5^2-3^2}}=-\frac{3}{4}$

$\pi <B<\frac{3\pi }{2}⟹\text{sec}B<0$

$\tan B=\frac{1}{2}⟹\text{sec}B=-\sqrt{1+\frac{1}{4}}=-\frac{\sqrt{5}}{2}$

So $8\tan A-\sqrt{5}\text{sec}B=8\times (-\frac{3}{4})-\sqrt{5}\times (-\frac{\sqrt{5}}{2})=-6+\frac{5}{2}=-\frac{7}{2}$