Question

If sin $A=\frac{3}{8}$ find the value of $1+11ta{n}^{2}A$

Answer 1

$\sin A=\frac{3}{8}$

$\Rightarrow \cos A=\sqrt{1-{\sin }^{2}A}$

$=\sqrt{1-\frac{9}{64}}$

$=\sqrt{\frac{64-9}{64}}$

$\Rightarrow \cos A=\frac{\sqrt{55}}{8},$ Now, $\tan A=\frac{\sin A}{\cos A}=\frac{3}{8}\times \frac{8}{\sqrt{55}}=\frac{3}{\sqrt{55}}$

$\Rightarrow {\tan }^{2}A=\frac{9}{55}\&1+11{\tan }^{2}A$

$=1+11\times \frac{3}{55}$

$=1+\frac{3}{5}$

$=\frac{8}{5}$

$\Rightarrow 1+11{\tan }^{2}A=\frac{8}{5}$

Hence, the answer is $\frac{8}{5}.$