Find the best tutors and institutes for Class 10 Tuition
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 $\text{cosec}^2 A = 1 + \cot^2 A$.
Solution :
Given: The trigonometric expression $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$.
To Prove: $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.
Step 1: Manipulating the Left-Hand Side (LHS)
We begin with the expression:
$LHS = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$
To express the terms in $\text{cosec } A$ and $\cot A$, we divide both the numerator and the denominator by $\sin A$.
$LHS = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}$
Step 2: Simplifying using Trigonometric Ratios
Recall the definitions: $\cot A = \frac{\cos A}{\sin A}$ and $\text{cosec } A = \frac{1}{\sin A}$.
Substituting these into the expression:
$LHS = \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A}$
Step 3: Applying the Trigonometric Identity
We are given the identity $\text{cosec}^2 A = 1 + \cot^2 A$. This can be rearranged as:
$1 = \text{cosec}^2 A - \cot^2 A$
We substitute this value of $1$ into the denominator of our expression:
$LHS = \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + (\text{cosec}^2 A - \cot^2 A)}$
Step 4: Factoring the Denominator
Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$, we factor $(\text{cosec}^2 A - \cot^2 A)$:
$LHS = \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$
To simplify, we factor out $(\cot A + \text{cosec } A)$ from the denominator. Note that $(\cot A - \text{cosec } A) = -(\text{cosec } A - \cot A)$:
$LHS = \frac{\cot A + \text{cosec } A - 1}{-(\text{cosec } A - \cot A) + (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$
Step 5: Final Cancellation
Factor out $(\text{cosec } A - \cot A)$ in the denominator:
$LHS = \frac{\cot A + \text{cosec } A - 1}{(\text{cosec } A - \cot A) [(\text{cosec } A + \cot A) - 1]}$
Since the term $(\cot A + \text{cosec } A - 1)$ appears in both the numerator and the denominator, they cancel out:
$LHS = \frac{1}{\text{cosec } A - \cot A}$
To reach the final form, multiply the numerator and denominator by the conjugate $(\text{cosec } A + \cot A)$:
$LHS = \frac{1 \cdot (\text{cosec } A + \cot A)}{(\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$
$LHS = \frac{\text{cosec } A + \cot A}{\text{cosec}^2 A - \cot^2 A}$
Since $\text{cosec}^2 A - \cot^2 A = 1$:
$LHS = \text{cosec } A + \cot A = RHS$
Final Answer: Since the Left-Hand Side equals the Right-Hand Side, the identity $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$ 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(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
Download free CBSE - Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.3 worksheets
Download Now
