# plane trigonometry

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james perez
plane trigonometry

can you help me solve ths? thank you

Jhun Vert

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

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

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

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

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

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

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

james perez

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

Orion1213

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.

Orion1213

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

james perez

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...

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?

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