Question

34. Prove that
cot theta/1+tan theta= cot theta - 1/ 2-sec²theta​

Answer 1

Step-by-step explanation:

Answer and Explanation:

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

1+tanθ

cotθ

=

2−sec

2

θ

cotθ−1

Solution :

Taking LHS,

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

1+tanθ

cotθ

Rationalize,

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

1+tanθ

cotθ

×

1−tanθ

1−tanθ

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

1

2

−tan

2

θ

cotθ−cotθtanθ

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

1−(sec

2

θ−1)

cotθ−1

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

1−sec

2

θ+1

cotθ−1

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

2−sec

2

θ

cotθ−1

=RHS

So, LHS=RHS