MP Board · Class 10 · Mathematics · Introduction to TrigonometryProve the trigonometric identity: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$. Also, state the fundamental trigonometric ratios in a right-angled triangle.

Introduction to Trigonometric Ratios\nTrigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. In a right-angled triangle, the primary trigonometric ratios with respect to an acute angle $\theta$ are defined as follows:

• Sine ($\sin \theta$): Ratio of the length of the side opposite to angle $\theta$ to the length of the hypotenuse. $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$
• Cosine ($\cos \theta$): Ratio of the length of the adjacent side to the length of the hypotenuse. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
• Tangent ($\tan \theta$): Ratio of the length of the side opposite to angle $\theta$ to the length of the adjacent side. $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

Proof of the Identity\nWe are required to prove the identity:

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

Step 1: Consider the Left Hand Side (LHS) $$\text{LHS} = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$$

Step 2: Take out the common factors from the numerator and the denominator

• In the numerator, $\sin \theta$ is common: $\sin \theta(1 - 2 \sin^2 \theta)$
• In the denominator, $\cos \theta$ is common: $\cos \theta(2 \cos^2 \theta - 1)$ \nSubstitute these back into the expression: $$\text{LHS} = \frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta - 1)}$$

Step 3: Apply fundamental trigonometric identities\nWe know that $\sin^2 \theta + \cos^2 \theta = 1$, which means $\sin^2 \theta = 1 - \cos^2 \theta$ and $\cos^2 \theta = 1 - \sin^2 \theta$.\nSubstitute $\sin^2 \theta = 1 - \cos^2 \theta$ in the numerator: $$1 - 2 \sin^2 \theta = 1 - 2(1 - \cos^2 \theta) = 1 - 2 + 2 \cos^2 \theta = 2 \cos^2 \theta - 1$$

Step 4: Simplify the expression\nNow, substitute this simplified numerator back into our fraction: $$\text{LHS} = \frac{\sin \theta (2 \cos^2 \theta - 1)}{\cos \theta (2 \cos^2 \theta - 1)}$$ \nThe term $(2 \cos^2 \theta - 1)$ cancels out from both the numerator and the denominator: $$\text{LHS} = \frac{\sin \theta}{\cos \theta}$$ \nSince $\frac{\sin \theta}{\cos \theta} = \tan \theta$, we get: $$\text{LHS} = \tan \theta = \text{RHS}$$ \nHence, the identity is successfully proved.