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Q1:
Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.
Solution :
Given: The trigonometric ratios $\sin A$, $\sec A$, and $\tan A$.
To Find: Express $\sin A$, $\sec A$, and $\tan A$ in terms of $\cot A$.
Step 1: Expressing $\tan A$ in terms of $\cot A$
By the definition of reciprocal trigonometric identities, we know that the tangent ratio is the reciprocal of the cotangent ratio.
Formula: $\tan A = \frac{1}{\cot A}$
Thus, $\tan A = \frac{1}{\cot A}$.
Step 2: Expressing $\sin A$ in terms of $\cot A$
We use the trigonometric identity relating $\csc A$ and $\cot A$: $1 + \cot^2 A = \csc^2 A$.
Taking the square root on both sides: $\csc A = \sqrt{1 + \cot^2 A}$.
Since $\sin A = \frac{1}{\csc A}$ [Reciprocal identity], we substitute the expression for $\csc A$:
$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$
Step 3: Expressing $\sec A$ in terms of $\cot A$
We use the trigonometric identity relating $\sec A$ and $\tan A$: $1 + \tan^2 A = \sec^2 A$.
Substitute the expression for $\tan A$ found in Step 1 ($\tan A = \frac{1}{\cot A}$):
$\sec^2 A = 1 + \left(\frac{1}{\cot A}\right)^2$
$\sec^2 A = 1 + \frac{1}{\cot^2 A}$
Find a common denominator:
$\sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A}$
Taking the square root on both sides:
$\sec A = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}}$
$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$
Final Answer:
$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$
$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$
$\tan A = \frac{1}{\cot A}$
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.3
- 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(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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