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Q4(ii):
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
Solution :
Given: An identity involving trigonometric ratios of an acute angle $A$: $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$.
To Prove: $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$.
Step 1: Taking the Left Hand Side (LHS) of the identity.
LHS = $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$
Step 2: Finding a common denominator.
To add the two fractions, we find the common denominator, which is $(1 + \sin A)(\cos A)$.
LHS = $\frac{(\cos A)(\cos A) + (1 + \sin A)(1 + \sin A)}{(1 + \sin A)(\cos A)}$
Step 3: Expanding the numerator.
LHS = $\frac{\cos^2 A + (1 + 2\sin A + \sin^2 A)}{(1 + \sin A)(\cos A)}$ [Using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ where $a=1, b=\sin A$]
Step 4: Applying the Pythagorean trigonometric identity.
We know that $\sin^2 A + \cos^2 A = 1$. Rearranging the terms in the numerator:
LHS = $\frac{(\sin^2 A + \cos^2 A) + 1 + 2\sin A}{(1 + \sin A)(\cos A)}$
LHS = $\frac{1 + 1 + 2\sin A}{(1 + \sin A)(\cos A)}$ [Substituting $\sin^2 A + \cos^2 A = 1$]
LHS = $\frac{2 + 2\sin A}{(1 + \sin A)(\cos A)}$
Step 5: Simplifying the expression.
Factor out the constant 2 from the numerator:
LHS = $\frac{2(1 + \sin A)}{(1 + \sin A)(\cos A)}$
Cancel the common term $(1 + \sin A)$ from the numerator and the denominator:
LHS = $\frac{2}{\cos A}$
Step 6: Converting to the final form.
Using the reciprocal identity $\frac{1}{\cos A} = \sec A$:
LHS = $2 \sec A$
Since LHS = RHS, the identity is proven.
Final Answer: The identity $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$ is successfully 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(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) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$
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