Question

HEY solve this one !!

Answer 1

1 = Sec²theta - tan²theta

Answer 2

               

$\sf{Using the ldentities} \ \ \sec^2\theta-\tan^2\theta=1\ \ \sf{\&} \ \ \sin\theta=\tan\theta \cdot \cos\theta,$

$\boxed{ LHS } \\ \\ \\ \boxed{\frac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}} \\ \\ \\ \boxed{\frac{(1+\sec\theta-\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}} \\ \\ \\ \boxed{\frac{(1+\sec\theta-\tan\theta)(\sec\theta-\tan\theta)}{\sec\theta-\tan\theta+\sec^2\theta-\tan^2\theta}}$

$\boxed{\frac{(1+\sec\theta-\tan\theta)(\sec\theta-\tan\theta)}{\sec\theta-\tan\theta+1}} \\ \\ \\ \boxed{\frac{(1+\sec\theta-\tan\theta)(\sec\theta-\tan\theta)}{1+\sec\theta-\tan\theta}} \\ \\ \\ \boxed{\sec\theta-\tan\theta} \\ \\ \\ \boxed{\frac{1}{\cos\theta}-\tan\theta} \\ \\ \\ \boxed{\frac{1-\tan\theta \cdot \cos\theta}{\cos\theta}} \\ \\ \\ \boxed{\frac{1-\sin\theta}{\cos\theta}} \\ \\ \\ \boxed{ RHS }$

$\sf{Hence proved! \\ \\ \\ Plz mark it as the brainliest. \\ \\ \\ Thank you. :-)}$