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Q4(vii):
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
Solution :
Given: An trigonometric expression involving an acute angle $\theta$: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$.
To Prove: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$.
Step 1: Analyze the Left-Hand Side (LHS)
The given expression is: $LHS = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$.
Step 2: Factor out common terms from the numerator and denominator
In the numerator, $\sin \theta$ is common to both terms. In the denominator, $\cos \theta$ is common to both terms.
Factoring the numerator: $\sin \theta - 2 \sin^3 \theta = \sin \theta (1 - 2 \sin^2 \theta)$.
Factoring the denominator: $2 \cos^3 \theta - \cos \theta = \cos \theta (2 \cos^2 \theta - 1)$.
Substituting these back into the LHS expression:
$LHS = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$
Step 3: Apply the Trigonometric Identity
Recall the fundamental Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$.
From this identity, we can express $1$ as $(\sin^2 \theta + \cos^2 \theta)$.
Substitute this into the numerator term $(1 - 2 \sin^2 \theta)$:
$1 - 2 \sin^2 \theta = (\sin^2 \theta + \cos^2 \theta) - 2 \sin^2 \theta$
$= \cos^2 \theta - \sin^2 \theta$
Now, substitute this into the denominator term $(2 \cos^2 \theta - 1)$:
$2 \cos^2 \theta - 1 = 2 \cos^2 \theta - (\sin^2 \theta + \cos^2 \theta)$
$= 2 \cos^2 \theta - \sin^2 \theta - \cos^2 \theta$
$= \cos^2 \theta - \sin^2 \theta$
Step 4: Simplify the expression
Substitute the simplified terms back into the LHS:
$LHS = \frac{\sin \theta (\cos^2 \theta - \sin^2 \theta)}{\cos \theta (\cos^2 \theta - \sin^2 \theta)}$
Since $\theta$ is an acute angle and the expression is defined, $(\cos^2 \theta - \sin^2 \theta) \neq 0$. Therefore, we can cancel the common factor $(\cos^2 \theta - \sin^2 \theta)$ from the numerator and denominator:
$LHS = \frac{\sin \theta}{\cos \theta}$
Step 5: Apply the Quotient Identity
Using the quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
$LHS = \tan \theta$
Conclusion:
Since $LHS = RHS$, the identity is proven.
Final Answer: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$ is proved.
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.3
- Q1: Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.
- Q2: Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.
- Q3(i): Choose the correct option. Justify your choice. (i) $9 \sec^2 A – 9 \tan^2 A =$
- Q3(ii): Choose the correct option. Justify your choice. (ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta – \text{cosec } \theta) =$
- Q3(iii): Choose the correct option. Justify your choice. (iii) $(\sec A + \tan A) (1 – \sin A) =$
- Q3(iv): Choose the correct option. Justify your choice. (iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$
- Q4(i): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) $(\text{cosec } \theta – \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
- Q4(ii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
- Q4(iii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$ [Hint : Write the expression in terms of $\sin \theta$ and $\cos \theta$]
- Q4(iv): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$ [Hint : Simplify LHS and RHS separately]
- Q4(ix): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ix) $(\text{cosec } A – \sin A) (\sec A – \cos A) = \frac{1}{\tan A + \cot A}$ [Hint : Simplify LHS and RHS separately]
- Q4(v): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.
- Q4(vi): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$
- Q4(viii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (viii) $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
- Q4(x): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (x) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$
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