# plane trigonometry

can you help me solve ths? thank you

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can you help me solve ths? thank you

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## Re: plane trigonometry

Prove the identity

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

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

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

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

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

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

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

## Re: plane trigonometry

In reply to Re: plane trigonometry 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

## Re: plane trigonometry

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.

## Re: plane trigonometry

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

## Re: plane trigonometry

In reply to Re: plane trigonometry 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...

## Re: plane trigonometry

In reply to Re: plane trigonometry by james perez

how to get then the length of the opposite side and adjacent f only the length of angle A,B,C were given?