Q.Show that $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$.

Recognise the triple-angle identity $4\cos^3\theta-3\cos\theta=\cos 3\theta$; here it gives $\cos 120°=-\dfrac{1}{2}$.

The multiple-angle formula for cosine is

$\cos 3\theta=4\cos^3\theta-3\cos\theta.$

With $\theta=40°$:

$4\cos^3 40°-3\cos 40°=\cos(3\times 40°)=\cos 120°.$

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$\sin15^\circ\cos15^\circ = \tfrac12\sin30^\circ = \tfrac12\cdot\tfrac12 = \tfrac14$, so the answer is option (c).

Recall the double-angle formula

$\sin 2\theta = 2\sin\theta\cos\theta \;\Rightarrow\; \sin\theta\cos\theta = \frac{1}{2}\sin 2\theta.$

With $\theta = 15^\circ$:

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Squaring $\sin A+\cos A=1$ and applying $\sin^2A+\cos^2A=1$ leaves $\sin 2A=0$.

Given:

$\sin A + \cos A = 1.$

Square both sides:

$(\sin A + \cos A)^2 = 1^2$

$\sin^2 A + 2\sin A\cos A + \cos^2 A = 1.$

Use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$: …