Question

If $cosecA+\sin A=5\frac{1}{5}$, the value of

$cose{c}^{2}A+{\sin }^{2}A$ is $m\frac{1}{25}$, m is

Answer 1

Given: $cosecA+\sin A=5\frac{1}{5}$

$\Rightarrow cosecA+\sin A=\frac{26}{5}$

Squaring both sides,

$\Rightarrow (cosecA+sinA{)}^2=(\frac{26}{5}{)}^2$

$\Rightarrow cose{c}^{2}A+si{n}^{2}A+2=\frac{676}{25}$

$\Rightarrow cose{c}^{2}A+si{n}^{2}A=\frac{626}{25}$

$\Rightarrow cose{c}^{2}A+si{n}^{2}A=25\frac{1}{25}$

Hence, $m=25$.