RBSE Solutions Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Exercise 14.2

Solve the Following by Using Trigonometric Identities (1 to 10)

Question 1. If cosec A = \( \frac {5}{4} \), then evaluate cot A, sin A and cos A.
Answer: We are given that \( \operatorname{cosec} A = \frac{5}{4} \).
We know the trigonometric identity: \( 1 + \cot^2 A = \operatorname{cosec}^2 A \).
\( \implies \cot^2 A = \operatorname{cosec}^2 A - 1 \)
\( \implies \cot A = \sqrt{\operatorname{cosec}^2 A - 1} \)
Now, substitute the value of \( \operatornamecosec A \):
\( = \sqrt{\left(\frac{5}{4}\right)^2 - 1} \)
\( = \sqrt{\frac{25}{16} - 1} \)
\( = \sqrt{\frac{25 - 16}{16}} \)
\( = \sqrt{\frac{9}{16}} \)
\( = \frac{3}{4} \)
So, \( \cot A = \frac{3}{4} \).
We also know that \( \operatorname{cosec} A = \frac{1}{\sin A} \). Since \( \operatornamecosec A = \frac{5}{4} \), it means \( \sin A = \frac{4}{5} \).
To find \( \cos A \), we use the identity \( \cos A = \sin A \times \cot A \).
\( = \frac{4}{5} \times \frac{3}{4} \)
\( = \frac{3}{5} \)
Therefore, the values are \( \cot A = \frac{3}{4} \), \( \sin A = \frac{4}{5} \), and \( \cos A = \frac{3}{5} \). These ratios help in understanding the angles in a right-angled triangle.
In simple words: First, we use a math rule to find 'cot A' from 'cosec A'. Then, we use another rule to find 'sin A' from 'cosec A'. Finally, we multiply 'sin A' and 'cot A' to find 'cos A'.

🎯 Exam Tip: Remember the fundamental trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \) to easily find other ratios.

Question 3. If sin A = \( \frac{3}{5} \), then evaluate cos A and tan A.
Answer: We are given that \( \sin A = \frac{3}{5} \).
We know the trigonometric identity: \( \cos^2 A = 1 - \sin^2 A \).
\( \implies \cos A = \sqrt{1 - \sin^2 A} \)
Now, substitute the given value of \( \sin A \):
\( = \sqrt{1 - \left(\frac{3}{5}\right)^2} \)
\( = \sqrt{1 - \frac{9}{25}} \)
\( = \sqrt{\frac{25 - 9}{25}} \)
\( = \sqrt{\frac{16}{25}} \)
\( = \frac{4}{5} \)
Thus, \( \cos A = \frac{4}{5} \). This is found directly from the sine value using the Pythagorean identity.
To find \( \tan A \), we use the formula \( \tan A = \frac{\sin A}{\cos A} \).
\( = \frac{3/5}{4/5} \)
\( = \frac{3}{4} \)
Therefore, \( \cos A = \frac{4}{5} \) and \( \tan A = \frac{3}{4} \).
In simple words: We find 'cos A' using a basic triangle rule with 'sin A'. Then, we divide 'sin A' by 'cos A' to get 'tan A'.

🎯 Exam Tip: Always remember the Pythagorean identities, especially \( \sin^2 A + \cos^2 A = 1 \), as they are very useful for finding one trigonometric ratio when another is known.

Question 4. If cos B = \( \frac {1}{3} \), then find the remaining trigonometrical ratios.
Answer: We are given that \( \cos B = \frac{1}{3} \).
First, let's find \( \sin B \). We know that \( \sin^2 B = 1 - \cos^2 B \).
\( \implies \sin B = \sqrt{1 - \cos^2 B} \)
Substitute the value of \( \cos B \):
\( = \sqrt{1 - \left(\frac{1}{3}\right)^2} \)
\( = \sqrt{1 - \frac{1}{9}} \)
\( = \sqrt{\frac{9 - 1}{9}} \)
\( = \sqrt{\frac{8}{9}} \)
\( = \frac{2\sqrt{2}}{3} \)
Next, let's find \( \tan B \). We use the formula \( \tan B = \frac{\sin B}{\cos B} \).
\( = \frac{2\sqrt{2}/3}{1/3} \)
\( = 2\sqrt{2} \)
Now, we find the reciprocal ratios:
\( \operatorname{cosec} B = \frac{1}{\sin B} = \frac{1}{2\sqrt{2}/3} = \frac{3}{2\sqrt{2}} \)
\( \sec B = \frac{1}{\cos B} = \frac{1}{1/3} = 3 \)
\( \cot B = \frac{1}{\tan B} = \frac{1}{2\sqrt{2}} \)
These six ratios fully describe the angles in a right-angled triangle.
In simple words: Starting with 'cos B', we first find 'sin B' using a basic rule. Then, we find 'tan B' by dividing 'sin B' by 'cos B'. After that, we just flip these three answers to get 'cosec B', 'sec B', and 'cot B'.

🎯 Exam Tip: Always list all six trigonometric ratios clearly. Remember their definitions and how they relate to each other (e.g., reciprocals) to avoid missing any part of the question.

Question 6. If tan A = \( \sqrt{2} - 1 \), then prove that sin A cos A = \( \frac {1}{2\surd 2} \).
Answer: Given that \( \tan A = \sqrt{2} - 1 \).
We can represent \( \tan A \) as \( \frac{\sqrt{2}-1}{1} \). In a right-angled triangle ABC, with angle A, let the opposite side (BC) be \( k(\sqrt{2}-1) \) and the adjacent side (AB) be \( k \), where \( k \) is a constant.
Using the Pythagorean theorem, \( AC^2 = AB^2 + BC^2 \):
\( AC = \sqrt{k^2 + \{k(\sqrt{2}-1)\}^2} \)
\( = \sqrt{k^2 + k^2( \sqrt{2}^2 - 2\sqrt{2} + 1 )} \)
\( = \sqrt{k^2 + k^2( 2 - 2\sqrt{2} + 1 )} \)
\( = \sqrt{k^2 + k^2( 3 - 2\sqrt{2} )} \)
\( = \sqrt{k^2( 1 + 3 - 2\sqrt{2} )} \)
\( = \sqrt{k^2( 4 - 2\sqrt{2} )} \)
\( = k\sqrt{4 - 2\sqrt{2}} \)
Now, let's find \( \sin A \) and \( \cos A \):
\( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{k(\sqrt{2}-1)}{k\sqrt{4-2\sqrt{2}}} = \frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}} \)
\( \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k}{k\sqrt{4-2\sqrt{2}}} = \frac{1}{\sqrt{4-2\sqrt{2}}} \)
Now we evaluate \( \sin A \cos A \):
\( \sin A \cos A = \left(\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}\right) \times \left(\frac{1}{\sqrt{4-2\sqrt{2}}}\right) \)
\( = \frac{\sqrt{2}-1}{4-2\sqrt{2}} \)
To simplify, we factor out common terms from the denominator:
\( = \frac{\sqrt{2}-1}{2(2-\sqrt{2})} \)
We notice that \( 2-\sqrt{2} = \sqrt{2}(\sqrt{2}-1) \).
\( = \frac{\sqrt{2}-1}{2\sqrt{2}(\sqrt{2}-1)} \)
Cancel out \( (\sqrt{2}-1) \) from numerator and denominator:
\( = \frac{1}{2\sqrt{2}} \)
This proves the statement. Understanding how to simplify expressions with radicals is key here.
In simple words: First, we draw a triangle using the given 'tan A' value. Then, we find the length of the third side using a special rule (Pythagoras). After that, we calculate 'sin A' and 'cos A' and multiply them. We simplify the answer to show it matches the number given in the question.

🎯 Exam Tip: When dealing with proofs involving trigonometric ratios, drawing a right-angled triangle and labeling its sides based on the given ratio can make the problem much clearer and easier to solve.

Question 7. If tan A = 2, then evaluate sec A . sin A + tan²A – cosec A.
Answer: We are given that \( \tan A = 2 \).
First, find \( \sec A \). We know the identity \( \sec^2 A = 1 + \tan^2 A \).
\( \implies \sec A = \sqrt{1 + \tan^2 A} \)
Substitute the value of \( \tan A \):
\( = \sqrt{1 + (2)^2} \)
\( = \sqrt{1 + 4} \)
\( = \sqrt{5} \)
Now, find \( \cos A \). Since \( \sec A = \frac{1}{\cos A} \), then \( \cos A = \frac{1}{\sec A} = \frac{1}{\sqrt{5}} \).
Next, find \( \sin A \). We know that \( \tan A = \frac{\sin A}{\cos A} \), so \( \sin A = \tan A \times \cos A \).
\( = 2 \times \frac{1}{\sqrt{5}} \)
\( = \frac{2}{\sqrt{5}} \)
Finally, find \( \operatorname{cosec} A \). Since \( \operatornamecosec A = \frac{1}{\sin A} \), then \( \operatorname{cosec} A = \frac{1}{2/\sqrt{5}} = \frac{\sqrt{5}}{2} \).
Now, we evaluate the expression: \( \sec A \cdot \sin A + \tan^2 A - \operatorname{cosec} A \).
Substitute all the calculated values:
\( = (\sqrt{5}) \cdot \left(\frac{2}{\sqrt{5}}\right) + (2)^2 - \left(\frac{\sqrt{5}}{2}\right) \)
\( = 2 + 4 - \frac{\sqrt{5}}{2} \)
\( = 6 - \frac{\sqrt{5}}{2} \)
To combine these, find a common denominator:
\( = \frac{12}{2} - \frac{\sqrt{5}}{2} \)
\( = \frac{12 - \sqrt{5}}{2} \)
This calculation shows how different trigonometric ratios combine into a single value.
In simple words: We are given 'tan A'. First, we find 'sec A' using a formula. Then, we find 'cos A' and 'sin A'. After that, we find 'cosec A'. Finally, we put all these values into the given expression and solve it step-by-step.

🎯 Exam Tip: When evaluating complex trigonometric expressions, always calculate each required ratio separately first. This prevents mistakes and keeps the main expression neat and easy to substitute into.

Question 9. If cos \( \theta = \frac {1}{\surd 2} \), then evaluate sin \( \theta \) and cot \( \theta \).
Answer: We are given that \( \cos \theta = \frac{1}{\sqrt{2}} \).
First, find \( \sin \theta \). We know the identity \( \sin^2 \theta = 1 - \cos^2 \theta \).
\( \implies \sin \theta = \sqrt{1 - \cos^2 \theta} \)
Substitute the given value of \( \cos \theta \):
\( = \sqrt{1 - \left(\frac{1}{\sqrt{2}}\right)^2} \)
\( = \sqrt{1 - \frac{1}{2}} \)
\( = \sqrt{\frac{2 - 1}{2}} \)
\( = \sqrt{\frac{1}{2}} \)
\( = \frac{1}{\sqrt{2}} \)
So, \( \sin \theta = \frac{1}{\sqrt{2}} \). This is a common value for a 45-degree angle.
Next, find \( \cot \theta \). We use the formula \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).
\( = \frac{1/\sqrt{2}}{1/\sqrt{2}} \)
\( = 1 \)
Therefore, \( \sin \theta = \frac{1}{\sqrt{2}} \) and \( \cot \theta = 1 \).
In simple words: We are given 'cos theta'. We use a math rule to find 'sin theta'. Since 'sin theta' and 'cos theta' are the same, 'cot theta' will be 1 because it's 'cos theta' divided by 'sin theta'.

🎯 Exam Tip: Recognize common trigonometric values like \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \) and \( \tan 45^\circ = \cot 45^\circ = 1 \). This can save time in calculations.

Question 10. If sec \( \theta = 2 \), then evaluate tan \( \theta \), cos \( \theta \) and sin \( \theta \).
Answer: We are given that \( \sec \theta = 2 \).
First, find \( \cos \theta \). We know that \( \cos \theta = \frac{1}{\sec \theta} \).
\( \implies \cos \theta = \frac{1}{2} \)
Next, find \( \tan \theta \). We know the identity \( \tan^2 \theta = \sec^2 \theta - 1 \).
\( \implies \tan \theta = \sqrt{\sec^2 \theta - 1} \)
Substitute the value of \( \sec \theta \):
\( = \sqrt{(2)^2 - 1} \)
\( = \sqrt{4 - 1} \)
\( = \sqrt{3} \)
So, \( \tan \theta = \sqrt{3} \). This is a standard value for \( \tan 60^\circ \).
Finally, find \( \sin \theta \). We know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), so \( \sin \theta = \tan \theta \times \cos \theta \).
\( = \sqrt{3} \times \frac{1}{2} \)
\( = \frac{\sqrt{3}}{2} \)
Therefore, \( \tan \theta = \sqrt{3} \), \( \cos \theta = \frac{1}{2} \), and \( \sin \theta = \frac{\sqrt{3}}{2} \).
In simple words: Given 'sec theta', we first find 'cos theta' by taking its inverse. Then, we use a formula to find 'tan theta'. Finally, we multiply 'tan theta' and 'cos theta' to get 'sin theta'.

🎯 Exam Tip: Always remember that \( \sec \theta \) is the reciprocal of \( \cos \theta \). Knowing this relationship is fundamental for solving problems efficiently.