Question

How to verify frac(sec theta/ csc theta- cot theta -frac(sectheta/csc theta +cot theta = 2csc theta

Answer 1

$\frac{sec \theta}{ csc \theta- cot \theta }-\frac{sec \theta}{csc \theta +cot \theta}= 2csc$


$\ sec(x)= \frac{1}{cos(x)} \\ csc(x)= \frac{1}{sin(x)} \\ cot(x)=\frac{sin(x)}{sin(x)} \\ sin^2(x)+cos^2(x)=1$


$\frac{sec \theta( csc \theta+cot \theta)}{ (csc \theta- cot \theta)( csc \theta+ cot \theta) }-\frac{sec \theta( csc \theta- cot \theta)}{( csc \theta- cot \theta)(csc \theta +cot \theta)} \\ =\frac{sec \theta csc \theta+sec \theta cot \theta-sec \theta csc \theta+sec \theta cot \theta}{ csc^2 \theta- cot^2 \theta }$
$=\frac{2sec \theta cot \theta}{ csc^2 \theta- cot^2 \theta } \\ = \frac{2 \times \frac{1}{cos\theta}\times\frac{cos\theta } {sin\theta }}{(\frac{1 } {sin\theta })^2-(\frac{cos\theta } {sin\theta })^2} \\ = \frac{\frac{2 } {sin\theta }}{\frac{1-cos^2\theta } {sin^2\theta }} \\ =\frac{\frac{2 } {sin\theta }}{\frac{sin^2\theta } {sin^2\theta }} \\ =\frac{2 } {sin\theta } \\ =2 \times \frac{1}{sin\theta } \\ =2csc \theta$


$\frac{sec \theta}{ csc \theta- cot \theta }-\frac{sec \theta}{csc \theta +cot \theta}= 2csc$