Question

Prove the following trigonometric identities:
tanθ-cotθ=2sin²θ-1/sinθcosθ

Answer 1

$\tan \theta-\cot \theta=\dfrac{2\sin^2 \theta-1}{\sin \theta\cos \theta}$, proved.

Step-by-step explanation:

To prove that the given trigonometric identities:

$\tan \theta-\cot \theta=\dfrac{2\sin^2 \theta-1}{\sin \theta\cos \theta}$

L.H.S. = $\tan \theta-\cot \theta$

Using the trigonometric identities,

$\tan \theta=\dfrac{\sin \theta}{\cos \theta}$ and $\cot \theta=\dfrac{\cos \theta}{\sin \theta}$

= $\dfrac{\sin \theta}{\cos \theta}-\dfrac{\cos \theta}{\sin \theta}$

Taking LCM of numerator part, we get

= $\dfrac{\sin^2 \theta-\cos^2 \theta}{\sin \theta\cos \theta}$

Using the trigonometric identity,

$\sin^2 \theta+\cos^2 \theta$ = 1

⇒ $\cos^2 \theta=1-\sin^2 \theta$ = 1

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

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

= $\dfrac{2\sin^2 \theta-1}{\sin \theta\cos \theta}$

= R.H.S., proved.

Thus, $\tan \theta-\cot \theta=\dfrac{2\sin^2 \theta-1}{\sin \theta\cos \theta}$, proved.