Question #9925d

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Question #9925d

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Answer 1 · 2017-01-10T02:42:16

$\frac{\frac{sin x}{cos x} - 1}{\frac{sin x}{cos x} + 1} = \frac{1 - \frac{cos x}{sin x}}{1 + \frac{cos x}{sin x}}$ $(sinx/cosx - 1)/(sinx/cosx + 1) = (1 - cosx/sinx)/(1 + cosx/sinx)$

$\frac{\frac{sin x - cos x}{cos x}}{\frac{sin x + cos x}{cos x}} = \frac{\frac{sin x - cos x}{sin x}}{\frac{sin x + cos x}{sin x}}$ $((sinx - cosx)/cosx)/((sinx + cosx)/cosx) = ((sinx - cosx)/sinx)/((sinx + cosx)/sinx)$

$\frac{sin x - cos x}{cos x} \cdot \frac{cos x}{sin x + cos x} = \frac{sin x - cos x}{sin x} \cdot \frac{sin x}{sin x + cos x}$ $(sinx - cosx)/cosx * cosx/(sinx + cosx) = (sinx - cosx)/sinx * sinx/(sinx + cosx)$

$\frac{sin x - cos x}{sin x + cos x} = \frac{sin x - cos x}{sin x + cos x}$ $(sinx - cosx)/(sinx + cosx) = (sinx - cosx)/(sinx + cosx)$

$L H S = R H S$ $LHS = RHS$

Identity proved!

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Question #9925d

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