Question

If $\tan \theta =\frac{a}{b}$, find ${\sin }^2\theta +{\cos }^2\theta$

Answer 1

$\tan \theta =\frac{a}{b}$

$\Rightarrow {\tan }^2\theta =\frac{a^2}{b^2}$

$\Rightarrow {\sec }^2\theta -1=\frac{a^2}{b^2}$

$\Rightarrow {\cos }^2\theta =\frac{b^2}{a^2+b^2}⟶\left(1\right)$

$\Rightarrow 1-{\sin }^2\theta =\frac{b^2}{a^2+b^2}$

$\Rightarrow {\sin }^2\theta =1-\frac{b^2}{a^2+b^2}=\frac{a^2+b^2-b^2}{a^2+b^2}$

$\Rightarrow {\sin }^2\theta =\frac{a^2}{a^2+b^2}$

$\therefore {\sin }^2\theta +{\cos }^2\theta =\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=\frac{a^2+b^2}{a^2+b^2}=1$

Hence, the answer is $1.$