Question

show that cot theta/1+tan theta= cot theta -1/2-sec.^2 theta

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer and Explanation:

To show : $\frac{\cot\theta}{1+\tan\theta}=\frac{\cot\theta-1}{2-\sec^2\theta}$

Solution :

Taking LHS,

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

Rationalize,

$=\frac{\cot\theta}{1+\tan\theta}\times\frac{1-\tan\theta}{1-\tan\theta}$

$=\frac{\cot\theta-\cot\theta\tan\theta}{1^2-\tan^2\theta}$

$=\frac{\cot\theta-1}{1-(\sec^2\theta-1)}$

$=\frac{\cot\theta-1}{1-\sec^2\theta+1}$

$=\frac{\cot\theta-1}{2-\sec^2\theta}$

=RHS

So, LHS=RHS