Question

If $\sin A=\frac{3}{5},\tan B=\frac{1}{2}$ and $\frac{\pi }{2}<A<\pi$ and $\pi <B<\frac{3\pi }{2}$ then $8\tan A-\sqrt{5}\sec B=$

• A. $-\frac{5}{2}$
• B. $\frac{5}{2}$
• C. $-\frac{7}{2}$
• D. $\frac{7}{2}$

Answer 1

The correct option is D $\frac{7}{2}$

Given,

$\sin A=\frac{3}{5}$

using Pythagoras theorem, we have,

$\Rightarrow \tan A=\frac{3}{4}$

$\tan B=\frac{1}{2}$

using Pythagoras theorem, we have,

$\Rightarrow \sec B=\frac{\sqrt{5}}{2}$

Given,

$8\tan A-\sqrt{5}\sec B$

$=8\times \frac{3}{4}-\sqrt{5}\frac{\sqrt{5}}{2}$

$=6-\frac{5}{2}$

$=\frac{7}{2}$