Question

2sinx/sin3x + tanx/tan3x =1 . prove this

Answer 1

Question #27434

2sinx/sin3x + tanx/tan3x = 1

we all know that

$$\sin 3x = 3 \sin x - 4 \sin^3 x \quad [1]$$ $$\cos 3x = 4 \cos^3 x - 3 \cos x \quad [2]$$ $$\tan x = \sin x / \cos x \quad [3]$$ $$\sin^2 x + \cos^2 x = 1 \quad [4]$$

substituting this into the problem statement

$$\frac{2 \sin x}{3 \sin x - 4 \sin^3 x} + \frac{\frac{\sin x}{\cos x}}{\frac{3 \sin x - 4 \sin^3 x}{4 \cos^3 x - 3 \cos x}} = \frac{2 \sin x}{3 \sin x - 4 \sin^3 x} + \frac{\sin x (4 \cos^3 x - 3 \cos x)}{\cos x (3 \sin x - 4 \sin^3 x)} = \frac{2 \sin x * \cos x + 4 \sin x * \cos^3 x - 3 \cos x * \sin x}{(3 \sin x - 4 \sin^3 x) * \cos x} = \frac{\cos x (2 \sin x + 4 \sin x * \cos^2 x - 3 \sin x)}{(3 \sin x - 4 \sin^3 x) * \cos x} = \frac{\sin x (4 \cos^2 x - 1)}{\sin x (3 - 4 \sin^2 x)}$$

Using formula [4]

$$4 \cos^2 x = 4 - 4 \sin^2 x$$

And substituting the result we get

$$\frac{4 - 4 \sin^2 x - 1}{3 - 4 \sin^2 x} = \frac{3 - 4 \sin^2 x}{3 - 4 \sin^2 x} = 1$$