Prove the given trigonometric equation — task. Maths CBSE Live product, Class 10 (2025-26).

Answer variants:

\frac{\sin θ ((\sin^{2} θ + \cos^{2} θ) + 2 \sin^{2} θ)}{\cos θ (2 \cos^{2} θ + (\sin^{2} θ + \cos^{2} θ))}

\frac{\sin θ (\sin^{2} θ + \cos^{2} θ - 2 \sin^{2} θ)}{\cos θ (2 \cos^{2} θ - \sin^{2} θ - \cos^{2} θ)}

\frac{\sin θ (1 + 2 \sin^{2} θ)}{\cos θ (2 \cos^{2} θ + 1)}

\frac{\tan θ (\cos^{2} θ - \sin^{2} θ)}{(\cos^{2} θ - \sin^{2} θ)}

\frac{\sin θ ((\sin^{2} θ + \cos^{2} θ) - 2 \sin^{2} θ)}{\cos θ (2 \cos^{2} θ - (\sin^{2} θ + \cos^{2} θ))}

\frac{\sin θ (1 - 2 \sin^{2} θ)}{\cos θ (2 \cos^{2} θ - 1)}

Prove that

\frac{\sin θ - 2 \sin^{3} θ}{2 \cos^{3} θ - \cos θ}

\(=\)

\tan θ

Proof:

LHS \(=\)

\frac{\sin θ - 2 \sin^{3} θ}{2 \cos^{3} θ - \cos θ}

\(=\)

\tan θ

\(=\) RHS

Did you find an error?

10. Prove the given trigonometric equation