Question

prove that sec ^2 theta -cot^2 (90-theta) = cos ^2 (90-theta) +cos ^2 theta

Answer 1

we solve LHS and RHS and
prove them equal

Answer 2

Answer:

Step-by-step explanation:

The given equation is :

$sec^{2}{\theta}-cot^{2}({90-\theta})=cos^{2}({90-\theta})+cos^{2}{\theta}}$

Taking the LHS of the above equation, we get

$sec^{2}{\theta}-cot^{2}({90-\theta})=sec^{2}{\theta}-tan^{2}{\theta}$ as ($tan^{2}{\theta}=cot^{2}({90-\theta})$)

=$1$

Taking the RHS of the given equation, we get

$cos^{2}({90-\theta})+cos^{2}{\theta}}$

=$sin^{2}({\theta})+cos^{2}{\theta}}$

=$1$

Since, LHS=RHS, hence proved.