Prove that $\dfrac{\sin \theta}{(\sec \theta - \tan \theta)(\sin \theta + 1)} = \tan \theta$ . Find all the values of $\theta$ between $0$ and $\pi$ for which $\dfrac{\sin \theta}{(\sec \theta - \tan \theta)(\sin \theta + 1)} = 1 - \sec^2 \theta$ .

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Worked solution

1. Express $\sec \theta$ and $\tan \theta$ in terms of $\sin \theta$ and $\cos \theta$ on the left-hand side (LHS).

$\text{LHS} = \frac{\sin \theta}{\left(\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}\right)(\sin \theta + 1)}$

2. Combine the terms in the first bracket and simplify the complex fraction.

\text{LHS} = \frac{\sin \theta}{\left(\frac{1 - \sin \theta}{\cos \theta}\right)(\sin \theta + 1)} = \frac{\sin \theta \cos \theta}{(1 - \sin \theta)(1 + \sin \theta)}

3. Expand the denominator using the difference of squares.

$\text{LHS} = \frac{\sin \theta \cos \theta}{1 - \sin^2 \theta}$

4. Apply the Pythagorean identity to the denominator and simplify to reach the right-hand side (RHS).

\text{LHS} = \frac{\sin \theta \cos \theta}{\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS}

Answer

Proof completed.

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