Question

Prove that ${\left(\sin \theta -sec\theta \right)}^2+{\left(\cos \theta -\cos ec\theta \right)}^2={\left(1-\sec \theta \cos ec\theta \right)}^2$

Answer 1

$LHS=\left({\sin }^2\theta -2\sin \theta \sec \theta +{\sec }^2\theta \right)+\left({\cos }^2\theta -2\cos \theta .\cos ec\theta +\cos e{c}^2\theta \right)$

$=\left({\sin }^2\theta +{\cos }^2\theta \right)+\left({\sec }^2\theta +\cos e{c}^2\theta \right)-2\left[\sin \theta .\sec \theta +\cos \theta .\cos ec\theta \right]$

$=1+\left[\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }\right]-2\left[\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }\right]$

$=1+\left[\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta .{\cos }^2\theta }\right]-2\left[\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta .\cos \theta }\right]$

$=1+\frac{1}{{\sin }^2\theta .{\cos }^2\theta }-\frac{2}{\sin \theta \cos \theta }$

$=1+{\sec }^2\theta .\cos e{c}^2\theta -2\sec \theta .\cos ec\theta$

$={\left(1-\sec \theta .\cos ec\theta \right)}^2$

LHS=RHS

Hence proof.