plane trigonometry


can you help me solve ths? thank you

Prove the identity
$\dfrac{2\sin x - 2\cos x}{\tan x - \cot x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)}{\dfrac{\sin x}{\cos x} - \dfrac{\cos x}{\sin x}} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)}{\dfrac{\sin^2 x- \cos^2 x}{\sin x \, \cos x}} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)\sin x \, \cos x}{\sin^2 x- \cos^2 x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)\sin x \, \cos x}{(\sin x - \cos x)(\sin x + \cos x)} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2\sin x \, \cos x}{\sin x + \cos x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{\sin 2x}{\sin x + \cos x} = \dfrac{\sin 2x}{\sin x + \cos x}$

In reply to by Jhun Vert

thank you sir.may i ask u another math problem? in right triangle , C= 90 degree, B= 60 degree, A= 30degree how will you find the length of the hypotenuse (c)? thanks sir

since the given triangle is a right triangle, that is, C=90 degrees... the sum of A + B = C... denotes it... hypotenuse can be solve, one of the legs, either side "a" (opposite to angle A) or side "b" (opposite to angle B) must be known... proceed solving hypotenuse "h" (opposite to angle C=90 deg)... and use either cosine or sine function.

better for you to solve it first... and show us your solution... that's the time we will help you...

In reply to by Orion1213

h= opposite side over 90 degree ( but the length of the opposite side is not given) that's my dilemma sir,either of the side is not given...