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Q4(i):
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) $(\text{cosec } \theta – \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
Solution :
Given: An identity involving trigonometric functions of an acute angle $\theta$, specifically $(\text{cosec } \theta - \cot \theta)^2$.
To Prove: $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
Step 1: Expressing the Left Hand Side (LHS) in terms of Sine and Cosine.
The given expression is $LHS = (\text{cosec } \theta - \cot \theta)^2$.
We use the fundamental trigonometric identities:
$\text{cosec } \theta = \frac{1}{\sin \theta}$
$\cot \theta = \frac{\cos \theta}{\sin \theta}$
Substituting these into the LHS:
$LHS = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2$
Step 2: Simplifying the fraction inside the square.
Since the denominators are the same, we combine the terms:
$LHS = \left( \frac{1 - \cos \theta}{\sin \theta} \right)^2$
$LHS = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$
Step 3: Applying the Pythagorean Identity.
We know the identity $\sin^2 \theta + \cos^2 \theta = 1$, which implies $\sin^2 \theta = 1 - \cos^2 \theta$.
Substituting this into the denominator:
$LHS = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$
Step 4: Factoring the denominator.
The denominator is in the form of a difference of squares, $a^2 - b^2 = (a - b)(a + b)$, where $a = 1$ and $b = \cos \theta$.
$1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$
Substituting this back into the expression:
$LHS = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}$
Step 5: Final Cancellation.
Canceling the common term $(1 - \cos \theta)$ from the numerator and the denominator:
$LHS = \frac{1 - \cos \theta}{1 + \cos \theta}$
Since $LHS = RHS$, the identity is proven.
Final Answer: Since the Left Hand Side simplifies to $\frac{1 - \cos \theta}{1 + \cos \theta}$, which is equal to the Right Hand Side, the identity $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$ is proven.
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(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(vii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
- 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) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
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