Question

Prove this identity sin^2θcsc^2θ=sin^2θ+cos^2θ by simplifying both sides of the equation.

Answer 1

$\bf \begin{array}{rllll} sin^2(\theta )csc^2(\theta )&=&sin^2(\theta )+cos^2(\theta )\\\\ ~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \cdot \cfrac{1^2}{~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }&=&1\\\\ 1^2&=&1\\ 1&=&1 \end{array}$

Answer 2

Answer:

Step-by-step explanation:

$sin^2 \theta csc^2 \theta=sin^2\theta+cos^2 \theta\\L.H.S.=sin ^2\theta cos^2\theta=\frac{sin^{2}\theta }{sin ^{2}\theta } =1\\R.H.S.=sin^{2} \theta+cos^{2} \theta=1\\L.H.S.=R.H.S$