Question

If $\tan A=\sqrt{2}-1$, then the value of $\csc A\cdot \sec A$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{2}{\sqrt{2}}$
• C. $2$
• D. $\frac{\sqrt{3}}{2}$
• E. none

Answer 1

The correct option is E none

REF.Image.

$tanA=\sqrt{2}-1$

$tanA=\frac{BC}{AB}=\frac{\sqrt{2}-1}{1}$

$BC=\sqrt{2}-1,AB=1$

$AC=\sqrt{4-2\sqrt{2}},sinA\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}},cosA=\frac{1}{\sqrt{4-2\sqrt{2}}}$

$cosecAsecA\Rightarrow \frac{\sqrt{4-2\sqrt{2}}}{\sqrt{2}-1}\times \frac{\sqrt{4-2\sqrt{2}}}{1}\Rightarrow \frac{4-2\sqrt{2}}{\sqrt{2}-1}$

$\Rightarrow \frac{2\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}-1)}=2\sqrt{2}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{4}{\sqrt{2}}$

Option (E) is correct.