Question

If $\csc A=\frac{7}{4}$, prove that $1+{\tan }^{2}A={\sec }^{2}A$.

Answer 1

$\csc A=\frac{7}{4}\left(\text{Given}\right)$

$\Rightarrow \sin A=\frac{4}{7}$

To prove:- $1+{\tan }^{2}A={\sec }^{2}A$

Proof:-

$1+{\tan }^{2}A={\sec }^{2}A$

$1+\frac{1}{{\cot }^{2}A}=\frac{1}{{\cos }^{2}A}$

$1+\frac{1}{{\csc }^{2}A-1}=\frac{1}{1-{\sin }^{2}A}$

$\Rightarrow 1+\frac{1}{\frac{1}{{\sin }^{2}A}-1}=\frac{1}{1-{\sin }^{2}A}$

$\Rightarrow 1+\frac{{\sin }^{2}A}{1-{\sin }^{2}A}=\frac{1}{1-{\sin }^{2}A}$

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

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

$\Rightarrow 1=1$

L.H.S. $=$ R.H.S.

Hence proved.