Prove that $\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta$.

Step 1: Factor out common terms

To simplify the given expression, we first factor out the common terms from the numerator and the denominator. In the numerator, $\sin \theta$ is common, and in the denominator, $\cos \theta$ is common.

Step 2: Apply trigonometric identities

We know the identity $\sin^2 \theta + \cos^2 \theta = 1$, which implies $\sin^2 \theta = 1 - \cos^2 \theta$. We substitute this into the numerator to express everything in terms of $\cos \theta$.

Step 3: Simplify the numerator

After substituting and simplifying the numerator, we get $1 - 2 + 2\cos^2 \theta = 2\cos^2 \theta - 1$. Now, the term $(2\cos^2 \theta - 1)$ appears in both the numerator and the denominator.

Step 4: Cancel common terms and conclude

Since $(2\cos^2 \theta - 1)$ is a common factor in both the numerator and the denominator, we can cancel it out. This leaves us with $\frac{\sin \theta}{\cos \theta}$, which is equal to $\tan \theta$ by definition.

More from this chapter

• **EXERCISE 8.3** **1.** Express the trigonometric ratios sin A, tan A and sec A in terms of cot A.
• Cos A / 1 + sin a + 1 + sin a / cos A is equal to 2 sec a
• write all the other trigonometric ratios of angle A in terms of sec A
• Express all the trigonometric ratios in terms of Sin A
• Express the trigonometric ratio sign a sec a tan a in terms of cot a
• Prove that √(sec² A + csc² A) = tan A + cot A
• (secθ - tanθ) = (1)/(m) --- (2) Adding (1) & (2) secθ + tanθ + secθ - tanθ = (1)/(m) + (m)/(1) 2secθ = (1 + m²)/(m) secθ = (1 + m²)/(2m) (Hence proved) 2) If as
• If tanθ + secθ = m, prove that secθ = (m² + 1)/(2m).

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