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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]
Solution :
Given: The trigonometric identity $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$, where $A$ is an acute angle.
To Prove: The Left Hand Side (LHS) is equal to the Right Hand Side (RHS).
Step 1: Simplifying the Left Hand Side (LHS)
The expression is given as: $LHS = \frac{1 + \sec A}{\sec A}$
We know the fundamental trigonometric identity relating secant and cosine: $\sec A = \frac{1}{\cos A}$ [Definition of secant as the reciprocal of cosine].
Substituting this into the LHS expression:
$LHS = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$
To simplify the numerator, find a common denominator:
$LHS = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}}$
Multiplying by the reciprocal of the denominator:
$LHS = \left( \frac{\cos A + 1}{\cos A} \right) \times \left( \frac{\cos A}{1} \right)$
Canceling the common term $\cos A$ in the numerator and denominator:
$LHS = 1 + \cos A$
Step 2: Simplifying the Right Hand Side (RHS)
The expression is given as: $RHS = \frac{\sin^2 A}{1 – \cos A}$
We use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$, which implies $\sin^2 A = 1 - \cos^2 A$ [Using the identity $\sin^2 \theta + \cos^2 \theta = 1$].
Substituting this into the RHS expression:
$RHS = \frac{1 - \cos^2 A}{1 - \cos A}$
Recognizing the numerator as a difference of squares, where $a^2 - b^2 = (a - b)(a + b)$. Here, $1 - \cos^2 A = (1)^2 - (\cos A)^2 = (1 - \cos A)(1 + \cos A)$:
$RHS = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A}$
Canceling the common term $(1 - \cos A)$ in the numerator and denominator (assuming $1 - \cos A \neq 0$, which is true for acute angle $A \neq 0^\circ$):
$RHS = 1 + \cos A$
Step 3: Conclusion
Since $LHS = 1 + \cos A$ and $RHS = 1 + \cos A$, we have shown that $LHS = RHS$.
Final Answer: Since both sides simplify to $1 + \cos A$, the identity $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$ 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(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(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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