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We are asked to prove two trigonometric identities. i. $\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x} = 2 \tan x \sec x$ ii. $\frac{\sin 2\theta}{\sin \theta} - \frac{\cos 2\theta}{\cos \theta} = \sec \theta$

TrigonometryTrigonometric IdentitiesDouble Angle FormulasTrigonometric Simplification
2025/3/31

1. Problem Description

We are asked to prove two trigonometric identities.
i. 11−sin⁡x−11+sin⁡x=2tan⁡xsec⁡x\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x} = 2 \tan x \sec x
ii. sin⁡2θsin⁡θ−cos⁡2θcos⁡θ=sec⁡θ\frac{\sin 2\theta}{\sin \theta} - \frac{\cos 2\theta}{\cos \theta} = \sec \theta

2. Solution Steps

i. To prove 11−sin⁡x−11+sin⁡x=2tan⁡xsec⁡x\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x} = 2 \tan x \sec x, we start with the left-hand side.
11−sin⁡x−11+sin⁡x=(1+sin⁡x)−(1−sin⁡x)(1−sin⁡x)(1+sin⁡x)=1+sin⁡x−1+sin⁡x1−sin⁡2x=2sin⁡xcos⁡2x\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x} = \frac{(1 + \sin x) - (1 - \sin x)}{(1 - \sin x)(1 + \sin x)} = \frac{1 + \sin x - 1 + \sin x}{1 - \sin^2 x} = \frac{2 \sin x}{\cos^2 x}
Using the identities tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} and sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}, we can rewrite the expression as:
2sin⁡xcos⁡2x=2⋅sin⁡xcos⁡x⋅1cos⁡x=2tan⁡xsec⁡x\frac{2 \sin x}{\cos^2 x} = 2 \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = 2 \tan x \sec x.
Thus, we have shown that the left-hand side equals the right-hand side, proving the identity.
ii. To verify sin⁡2θsin⁡θ−cos⁡2θcos⁡θ=sec⁡θ\frac{\sin 2\theta}{\sin \theta} - \frac{\cos 2\theta}{\cos \theta} = \sec \theta, we start with the left-hand side.
Using the double angle formulas:
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2 \theta - \sin^2 \theta
So,
sin⁡2θsin⁡θ−cos⁡2θcos⁡θ=2sin⁡θcos⁡θsin⁡θ−cos⁡2θ−sin⁡2θcos⁡θ=2cos⁡θ−cos⁡2θcos⁡θ+sin⁡2θcos⁡θ=2cos⁡θ−cos⁡θ+sin⁡2θcos⁡θ=cos⁡θ+sin⁡2θcos⁡θ\frac{\sin 2\theta}{\sin \theta} - \frac{\cos 2\theta}{\cos \theta} = \frac{2 \sin \theta \cos \theta}{\sin \theta} - \frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta} = 2 \cos \theta - \frac{\cos^2 \theta}{\cos \theta} + \frac{\sin^2 \theta}{\cos \theta} = 2 \cos \theta - \cos \theta + \frac{\sin^2 \theta}{\cos \theta} = \cos \theta + \frac{\sin^2 \theta}{\cos \theta}
=cos⁡2θ+sin⁡2θcos⁡θ=1cos⁡θ=sec⁡θ= \frac{\cos^2 \theta + \sin^2 \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta.
Thus, we have shown that the left-hand side equals the right-hand side, verifying the identity.

3. Final Answer

i. 11−sin⁡x−11+sin⁡x=2tan⁡xsec⁡x\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x} = 2 \tan x \sec x
ii. sin⁡2θsin⁡θ−cos⁡2θcos⁡θ=sec⁡θ\frac{\sin 2\theta}{\sin \theta} - \frac{\cos 2\theta}{\cos \theta} = \sec \theta

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