Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{\text{(sinθ - 2sin³θ)}}{\text{(2cos³θ - cosθ)}} = \text{tan θ}\)
Solution and Explanation
\(\frac{\text{(sinθ - 2sin³θ)}}{\text{(2cos³θ - cosθ)}} = \text{tan θ}\)
L.H.S =\(\frac{\text{(sinθ - 2sin³θ)}}{\text{(2cos³θ - cosθ)}}\)
Simplifying the L.H.S:
\(\frac{\text{sinθ (1 - 2sin²θ)}}{\text{cosθ {2(1 - sin²θ) - 1}}}\)
\(= \frac{\text{sinθ (1 - 2sin²θ)}}{\text{cosθ(2 - 2sin²θ - 1)}}\)
\(=\frac{\text{ sinθ (1 - 2sin²θ)}}{\text{cosθ(1 - 2sin²θ)}}\)
\(=\frac{ \text{sinθ}}{\text{cosθ}}\)
\(=\text{ tanθ}\)
= R.H.S
Top Questions on Trigonometric Identities
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.
- JEE Main - 2026
- Mathematics
- Trigonometric Identities
If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to
- JEE Main - 2026
- Mathematics
- Trigonometric Identities
The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
- JEE Main - 2026
- Mathematics
- Trigonometric Identities
- Prove the identity \( \sec^2 \theta = 1 + \tan^2 \theta \) for any right-angled triangle and use it to show that: \[ \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}. \]
- UK Class X - 2026
- Mathematics
- Trigonometric Identities
- Want to practice more? Try solving extra ecology questions todayView All Questions
