Question

prove that
sec^2 theta+cosec^2 (180°-theta)=sec^2theta
.cosec^2theta ​

Answer 1

Answer:

To prove:-

$\sf sec^2\theta+cosec^2(180°- \theta)=sec^2 \theta.cosec^2\theta$

Proof:-

LHS:-

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow sec^2\theta+cosec^2 (180°-\theta)$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow sec^2\theta+\left\{ cosec (180°- \theta)\right\}^2$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow sec^2\theta+cosec^2\theta$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac{1}{cos^2\theta}+\dfrac {1}{sin^2\theta}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {sin^2\theta+cos^2\theta}{sin^2\theta.cos^2\theta}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac{1}{sin^2\theta.cos^2\theta}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {1}{sin^2\theta}.\dfrac {1}{cos^2\theta}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow cosec^2\theta.sec^2\theta$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow RHS$

$\\\\\therefore\sf LHS=RHS\qquad_{(Proved)}$