Question

Prove the following trigonometric identities:
1+cot²θ/1+cosecθ=cosecθ

Answer 1

The given trigonometric identity, $1+\dfrac{\cot^2 \theta}{1+\csc \theta} =\csc \theta$, proved.

Step-by-step explanation:

To prove that the given trigonometric identity:

$1+\dfrac{\cot^2 \theta}{1+\csc \theta} =\csc \theta$.

L.H.S. = $1+\dfrac{\cot^2 \theta}{1+\csc \theta}$

Taking LCM of denominator part, we get

= $\dfrac{1+\csc \theta+\cot^2 \theta}{1+\csc \theta}$

Using the trigonometric identity,

$\csc^2 A-\cot^2 A$ = 1

⇒ $\cot^2 A=\csc^2 A-1$

= $\dfrac{1+\csc \theta+\csc^2 \theta-1}{1+\csc \theta}$

= $\dfrac{\csc \theta+\csc^2 \theta}{1+\csc \theta}$

Taking $\csc \theta$ common as numerator, we get

= $\dfrac{\csc \theta(1+\csc \theta)}{(1+\csc \theta)}$

= $\csc \theta$

= R.H.S., proved.

Thus, the given trigonometric identity, $1+\dfrac{\cot^2 \theta}{1+\csc \theta} =\csc \theta$, proved.

Answer 2

Answer:

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