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Q7(i):
If $\cot \theta = \frac{7}{8}$, evaluate : (i) $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$
Solution :
Given: $\cot \theta = \frac{7}{8}$
To find: The value of $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$
Step 1: Define the trigonometric ratio in terms of sides.
In a right-angled triangle, $\cot \theta = \frac{\text{Adjacent side}}{\text{Opposite side}}$.
Given $\cot \theta = \frac{7}{8}$, let the adjacent side $AB = 7k$ and the opposite side $BC = 8k$, where $k$ is a positive constant.
Step 2: Calculate the hypotenuse using the Pythagorean Theorem.
According to the Pythagorean Theorem: $AC^2 = AB^2 + BC^2$.
$AC^2 = (7k)^2 + (8k)^2$
$AC^2 = 49k^2 + 64k^2$
$AC^2 = 113k^2$
$AC = \sqrt{113}k$
Step 3: Determine $\sin \theta$ and $\cos \theta$.
$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}$
$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}$
Step 4: Simplify the expression using algebraic identities.
The expression is $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$.
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$:
Numerator: $(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta$
Denominator: $(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta$
Expression = $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta}$
Step 5: Substitute the values and calculate.
$1 - \sin^2 \theta = 1 - (\frac{8}{\sqrt{113}})^2 = 1 - \frac{64}{113} = \frac{113 - 64}{113} = \frac{49}{113}$
$1 - \cos^2 \theta = 1 - (\frac{7}{\sqrt{113}})^2 = 1 - \frac{49}{113} = \frac{113 - 49}{113} = \frac{64}{113}$
Expression = $\frac{49/113}{64/113} = \frac{49}{113} \times \frac{113}{64} = \frac{49}{64}$
Final Answer: \frac{49}{64}
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1
- Q1(i): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (i) $\sin A, \cos A$
- Q1(ii): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (ii) $\sin C, \cos C$
- Q10: In $\triangle PQR$, right-angled at $Q$, $PR + QR = 25$ cm and $PQ = 5$ cm. Determine the values of $\sin P, \cos P$ and $\tan P$.
- Q11(i): State whether the following are true or false. Justify your answer. (i) The value of $\tan A$ is always less than 1.
- Q11(ii): State whether the following are true or false. Justify your answer. (ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.
- Q11(iii): State whether the following are true or false. Justify your answer. (iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.
- Q11(iv): State whether the following are true or false. Justify your answer. (iv) $\cot A$ is the product of cot and $A$.
- Q11(v): State whether the following are true or false. Justify your answer. (v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.
- Q2: In Fig. 8.13, find $\tan P – \cot R$.
- Q3: If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.
- Q4: Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.
- Q5: Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.
- Q6: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.
- Q7(ii): If $\cot \theta = \frac{7}{8}$, evaluate : (ii) $\cot^2 \theta$
- Q8: If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A$ or not.
- Q9(i): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (i) $\sin A \cos C + \cos A \sin C$
- Q9(ii): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (ii) $\cos A \cos C – \sin A \sin C$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
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