Question

Prove that (cosectheta-sintheta)(sectheta-cos theta)=1/tantheta+cottheta

Answer 1

your answer is explained above

Answer 2

Answer and explanation:

To prove : $(csc\theta-\sin\theta)(\sec\theta-\cos\theta)=\frac{1}{\tan\theta+\cot\theta}$

Proof :

Taking LHS,

$LHS=(csc\theta-\sin\theta)(\sec\theta-\cos\theta)$

$LHS=(\frac{1}{\sin\theta}-\sin\theta)(\frac{1}{\cos\theta}-\cos\theta)$

$LHS=(\frac{1-\sin^2\theta}{\sin\theta})(\frac{1-\cos^2\theta}{\cos\theta})$

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

$LHS=\sin\theta\cos\theta$

Multiply and divide by $\cos\theta$

$LHS=\tan\theta\cos^2\theta$

$LHS=\frac{\tan\theta}{\sec^2\theta}$

$LHS=\frac{\tan\theta}{1+\tan^2\theta}$

$LHS=\frac{1}{\frac{1}{\tan\theta}+\frac{\tan^2\theta}{\tan\theta}}$

$LHS=\frac{1}{\cot\theta+\tan\theta}$

$LHS=RHS$

Hence proved.