Question

If Sin theta = 2mn/(m^2 + n^2), then show that: (Sin theta * Cot theta)/Cos theta = 1
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Answer 1

$\dfrac{\sin \theta.\cot \theta}{\cos \theta}= 1$, proved.

Step-by-step explanation:

We have,

$\sin \theta=\dfrac{2mn}{m^2 + n^2}$

Show that, $\dfrac{\sin \theta.\cot \theta}{\cos \theta}= 1$]

L.H.S.$=\dfrac{\sin \theta.\cot \theta}{\cos \theta}$

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

Using trigonometric identity,

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

= 1

= R.H.S.

Hence, $\dfrac{\sin \theta.\cot \theta}{\cos \theta}= 1$, proved.