Question

tan theta minus cot theta equal to sin square theta minus one upon sin theta cos theta​

Answer 1

Step-by-step explanation:

is that your question?

Answer 2

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

Step-by-step explanation:

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

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

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

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

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

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

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

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

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

= R.H.S. , proved.

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