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Prove that: tan5x+tan3x/tan5x-tan3x=4cos2xcos4x

Prove that: tan5x+tan3x/tan5x-tan3x=4cos2xcos4x

Grade:12

2 Answers

Latika Leekha
askIITians Faculty 165 Points
6 years ago
Hello student,
We first consider the L.H.S. and try to reach the R.H.S.
L.H.S = \frac{tan 5x + tan 3x}{tan 5x - tan 3x}
= \frac{\frac{sin 5x}{cos 5x}+\frac{sin 3x}{cos 3x}}{\frac{sin 5x}{cos 5x}-\frac{sin 3x}{cos 3x}}
= \frac{sin 5x cos 3x +sin 3x cos 5x}{sin 5x cos 3x - sin 3x cos 5x}
= \frac{sin (5x+3x)}{sin (5x-3x)}
(Using the formula sin A cos B + cos A sin B = sin (A+B) )
= \frac{sin 8x}{sin 2x}
= 2 sin 4x cos 4x/sin 2x
= 2(2 sin 2x cos 2x)cos 4x/sin 2x
= 4 cos 2x cos 4x.
= R.H.
Sai Sampath
34 Points
6 years ago
=\frac{sin 5x cos 3x +sin 3x cos 5x}{sin 5x cos 3x - sin 3x cos 5x}= \frac{\frac{sin 5x}{cos 5x}+\frac{sin 3x}{cos 3x}}{\frac{sin 5x}{cos 5x}-\frac{sin 3x}{cos 3x}}= \frac{tan 5x + tan 3x}{tan 5x - tan 3x}(Using the formula sin A cos B + cos A sin B = sin (A+B)
\frac{sin (5x+3x)}{sin (5x-3x)} 
 = 2 sin 4x cos 4x/sin 2x= 2(2 sin 2x cos 2x)cos 4x/sin 2x= 4 cos 2x cos 4x.\frac{sin 8x}{sin 2x} 
 
 

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