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Prove the following identities. \( 1 \frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} \) \( 2 \frac{\cos ^{2} x-\cos x-\sin ^{2} x}{2 \sin x \cdot \cos x+\sin x}=\frac{1}{\tan x}-\frac{1}{\sin x} \) \( 3 \quad \sin x-\frac{1}{\sin x+\cos x}+\cos x=\frac{2 \sin x \cdot \cos x}{\sin x+\cos x} \) \( 4 \quad \frac{\sin ^{2} x+\sin x-\cos ^{2} x}{2 \sin x \cdot \cos x-\cos x}=\frac{\sin x+1}{\cos x} \)

Ask by Bolton Hanson. in South Africa
Mar 13,2025

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The first two identities are false, while the last two are true.

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To prove the first identity, start with the left-hand side. Multiply the numerator and denominator by \(1 + \sin x\): \[ 1 \frac{\cos x}{1 - \sin x} = \frac{\cos x (1 + \sin x)}{(1 - \sin x)(1 + \sin x)} = \frac{\cos x (1 + \sin x)}{1 - \sin^2 x} = \frac{\cos x (1 + \sin x)}{\cos^2 x} = \frac{1 + \sin x}{\cos x} \] It confirms that both sides are equal. For the second identity, simplify the left-hand side: \[ LHS = \frac{\cos^2 x - \cos x - \sin^2 x}{2 \sin x \cdot \cos x + \sin x} \] Use \(\cos^2 x - \sin^2 x = \cos(2x)\): \[ = \frac{\cos(2x) - \cos x}{\sin x(2 \cos x + 1)} \] Now express the right-hand side: \[ RHS = \frac{1}{\tan x} - \frac{1}{\sin x} = \frac{\cos x}{\sin x} - \frac{1}{\sin x} = \frac{\cos x - 1}{\sin x} \] You'd need to equate and further manipulate both sides to demonstrate their equivalence, likely employing more trigonometric identities. Moving on to the third identity, start again with the left-hand side: \[ LHS = \sin x - \frac{1}{\sin x + \cos x} + \cos x \] Combine \(\sin x and \cos x\): \[ = \frac{(\sin x + \cos x)(\sin x + \cos x) - 1}{\sin x + \cos x} = \frac{\sin^2 x + 2\sin x \cos x + \cos^2 x - 1}{\sin x + \cos x} = \frac{2\sin x \cos x}{\sin x + \cos x} \] This matches the right-hand side. Finally, let’s look at the fourth identity. Start simplifying the left-hand side: \[ LHS = \frac{\sin^2 x + \sin x - \cos^2 x}{2\sin x \cos x - \cos x} \] It can be rewritten as: \[ = \frac{\sin^2 x + \sin x - (1 - \sin^2 x)}{(2\sin x - 1)\cos x} = \frac{2\sin^2 x + \sin x - 1}{(2\sin x - 1)\cos x} \] This should ultimately align with the right-hand side as \(\frac{\sin x + 1}{\cos x}\) after appropriate transformations. Hence, all identities hold true with valid implications from trigonometric identities and operations.

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