Question

how to do this question? Please tell me the answer with steps

Answer 1

$\dfrac{2\cos \theta \sin \theta}{\tan(90-\theta)}=2\sin^2 \theta\\\\ \dfrac{2\cos \theta \sin \theta}{\cot \theta}=2\sin^2 \theta\\\\ \dfrac{2\cos \theta \sin \theta}{\dfrac{\cos \theta}{\sin \theta}}=2\sin^2 \theta\\\\ 2\cos \theta \sin \theta\cdot\dfrac{\sin \theta}{\cos \theta}=2\sin^2 \theta\\\\ 2\sin^2 \theta=2\sin^2 \theta$

Answer 2

Consider a right angled triangle with hypotenuse c and legs  a and b. Let sin theta =  a/c .

Then cos theta =  b/c  and  tan (90-theta) = b/a

Then  2 sin theta cos theta /  tan (90 - theta)

= 2 * a/c * b/c  /  b/a

= 2 *  ab/c^2 * a/b

= 2 a^2 /  c^2   = 2 sin^2 theta