Question

$se{c}^2\left(ta{n}^{-1}\right(2\left)\right)+cose{c}^2\left(co{t}^{-1}\right(3\left)\right)=15$

Answer 1

$se{c}^2\left(ta{n}^{-1}\right(2\left)\right)+cose{c}^2\left(co{t}^{-1}\right(3\left)\right)=15$
Let
$\begin{array}{cc}ta{n}^{-1}2=\theta & co{t}^{-1}3=\theta \\ \Rightarrow sec\theta =\sqrt{5}& cot\theta =3\\ \therefore se{c}^2\left(se{c}^{-1}\sqrt{5}\right)& \therefore cosec\theta =\sqrt{10}\\ =5& cose{c}^2\left(co{t}^{-1}3\right)=10\end{array}$
$\therefore se{c}^2\left[ta{n}^{-1}2\right]+cose{c}^2\left[co{t}^{-1}3\right]=15$