Samacheer Kalvi Class 9 Maths Solutions Chapter 6 Trigonometry Exercise 6.5

Question 1. If \( \sin 30^\circ = x \) and \( \cos 60^\circ = y \), then \( x^2 + y^2 \) is .......
(a) \( \frac{1}{2} \)
(b) 0
(c) \( \sin 90^\circ \)
(d) \( \cos 90^\circ \)
Answer: (a) \( \frac{1}{2} \)
\( x = \sin 30^\circ = \frac{1}{2} \)
\( y = \cos 60^\circ = \frac{1}{2} \)
Now, we need to find \( x^2 + y^2 \).
\( x^2 + y^2 = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 \)
\( = \frac{1}{4} + \frac{1}{4} \)
\( = \frac{2}{4} \)
\( = \frac{1}{2} \)
So, the sum of the squares of x and y is \( \frac{1}{2} \). Trigonometric values like \( \sin 30^\circ \) and \( \cos 60^\circ \) are common in geometry and physics problems.
In simple words: First, find the values of x and y using the given angles. Then, square both x and y and add the results together.

🎯 Exam Tip: Remember the standard trigonometric values for common angles like 0°, 30°, 45°, 60°, and 90°. This helps solve problems quickly.

Question 2. If \( \tan \theta = \cot 37^\circ \), then the value of \( \theta \) is .........
(a) \( 37^\circ \)
(b) \( 53^\circ \)
(c) \( 90^\circ \)
(d) \( 1^\circ \)
Answer: (b) \( 53^\circ \)
Given that \( \tan \theta = \cot 37^\circ \).
We know that \( \cot A = \tan (90^\circ - A) \).
So, \( \cot 37^\circ = \tan (90^\circ - 37^\circ) \)
\( = \tan 53^\circ \)
Therefore, \( \tan \theta = \tan 53^\circ \).
This means \( \theta = 53^\circ \). This property of complementary angles is very useful in simplifying trigonometric expressions.
In simple words: Since \( \tan \theta \) equals \( \cot 37^\circ \), and \( \cot 37^\circ \) is the same as \( \tan (90^\circ - 37^\circ) \), then \( \theta \) must be \( 53^\circ \).

🎯 Exam Tip: Always remember the complementary angle identities, such as \( \tan \theta = \cot (90^\circ - \theta) \) and \( \sin \theta = \cos (90^\circ - \theta) \), as they are key to solving many trigonometry problems.

Question 3. The value of \( \tan 72^\circ \cdot \tan 18^\circ \) is .........
(a) 0
(b) 1
(c) \( 18^\circ \)
Answer: (b) 1
We need to find the value of \( \tan 72^\circ \cdot \tan 18^\circ \).
We know that \( \tan \theta = \cot (90^\circ - \theta) \).
So, \( \tan 72^\circ = \cot (90^\circ - 72^\circ) \)
\( = \cot 18^\circ \)
Now, substitute this back into the expression:
\( \tan 72^\circ \cdot \tan 18^\circ = \cot 18^\circ \cdot \tan 18^\circ \)
We also know that \( \cot A = \frac{1}{\tan A} \).
So, \( \cot 18^\circ \cdot \tan 18^\circ = \frac{1}{\tan 18^\circ} \cdot \tan 18^\circ \)
\( = 1 \). This identity shows how angles that add up to 90 degrees relate in tangent and cotangent functions.
In simple words: Since \( 72^\circ \) and \( 18^\circ \) add up to \( 90^\circ \), \( \tan 72^\circ \) is the same as \( \cot 18^\circ \). When you multiply \( \cot 18^\circ \) by \( \tan 18^\circ \), the answer is always 1.

🎯 Exam Tip: When you see angles that add up to \( 90^\circ \) (complementary angles) in tangent/cotangent problems, always look to use the identity \( \tan \theta \cdot \tan (90^\circ - \theta) = 1 \).

Question 4. The value of \( \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} \) is equal to .........
(a) \( \cos 60^\circ \)
(b) \( \sin 60^\circ \)
(c) \( \tan 60^\circ \)
(d) \( \sin 30^\circ \)
Answer: (c) \( \tan 60^\circ \)
We need to evaluate the expression \( \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} \).
First, we know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Now, substitute this value into the expression:
\( \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 - \left( \frac{1}{\sqrt{3}} \right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{3 - 1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \)
To simplify, we multiply the numerator by the reciprocal of the denominator:
\( = \frac{2}{\sqrt{3}} \times \frac{3}{2} \)
\( = \frac{3}{\sqrt{3}} \)
We can rationalize the denominator by multiplying the numerator and denominator by \( \sqrt{3} \):
\( = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{3} \)
\( = \sqrt{3} \)
We also know that \( \tan 60^\circ = \sqrt{3} \). This expression is also a standard double angle formula for tangent, \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \).
In simple words: Replace \( \tan 30^\circ \) with its value. Then, do the math carefully. The final answer, \( \sqrt{3} \), is the same as \( \tan 60^\circ \).

🎯 Exam Tip: This is the formula for \( \tan 2A \). Recognizing such formulas can save a lot of calculation time. Always remember the common trigonometric values.

Question 5. If \( 2 \sin 2\theta = \sqrt{3} \) then the value of \( \theta \) is .........
(a) \( 90^\circ \)
(b) \( 30^\circ \)
(c) \( 45^\circ \)
(d) \( 60^\circ \)
Answer: (b) \( 30^\circ \)
Given the equation \( 2 \sin 2\theta = \sqrt{3} \).
Divide both sides by 2:
\( \sin 2\theta = \frac{\sqrt{3}}{2} \)
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
So, we can write:
\( \sin 2\theta = \sin 60^\circ \)
This means that \( 2\theta = 60^\circ \).
Now, divide by 2 to find \( \theta \):
\( \theta = \frac{60^\circ}{2} \)
\( \implies \theta = 30^\circ \). Knowing these standard sine values helps in solving many trigonometric equations easily.
In simple words: First, find what \( \sin 2\theta \) equals. Then, find which angle has that sine value. Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), then \( 2\theta \) is \( 60^\circ \), which means \( \theta \) is \( 30^\circ \).

🎯 Exam Tip: When solving trigonometric equations, always isolate the trigonometric function first (like \( \sin 2\theta \)), then find the angle. Be careful if the angle itself is a multiple, like \( 2\theta \).

Question 6. The value of \( 3 \sin 70^\circ \sec 20^\circ + 2 \sin 39^\circ \sec 51^\circ \) is .........
(a) 2
(b) 3
(c) 5
(d) 6
Answer: (c) 5
We need to find the value of \( 3 \sin 70^\circ \sec 20^\circ + 2 \sin 39^\circ \sec 51^\circ \).
Let's use the complementary angle identities:
\( \sec A = \cosec (90^\circ - A) \)
Also, \( \sec (90^\circ - A) = \cosec A \).
For the first term, \( \sec 20^\circ = \sec (90^\circ - 70^\circ) = \cosec 70^\circ \).
For the second term, \( \sec 51^\circ = \sec (90^\circ - 39^\circ) = \cosec 39^\circ \).
Now substitute these into the expression:
\( 3 \sin 70^\circ (\cosec 70^\circ) + 2 \sin 39^\circ (\cosec 39^\circ) \)
We know that \( \cosec A = \frac{1}{\sin A} \).
So, \( \sin 70^\circ \cdot \cosec 70^\circ = \sin 70^\circ \cdot \frac{1}{\sin 70^\circ} = 1 \).
And \( \sin 39^\circ \cdot \cosec 39^\circ = \sin 39^\circ \cdot \frac{1}{\sin 39^\circ} = 1 \).
Therefore, the expression becomes:
\( 3(1) + 2(1) \)
\( = 3 + 2 \)
\( = 5 \). This calculation demonstrates the power of trigonometric identities in simplifying complex expressions.
In simple words: We change \( \sec 20^\circ \) to \( \cosec 70^\circ \) and \( \sec 51^\circ \) to \( \cosec 39^\circ \). Since \( \sin \theta \) multiplied by \( \cosec \theta \) always equals 1, the problem simplifies to \( 3 \times 1 + 2 \times 1 \), which is 5.

🎯 Exam Tip: Always look for angles that are complementary (add up to \( 90^\circ \)) when sine and cosine, or secant and cosecant, or tangent and cotangent, are involved. This often leads to significant simplifications.

Question 7. The value of \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} \) is ........
(a) 2
(c) 0
(d) \( \frac{1}{2} \)
Answer: (c) 0
We need to find the value of \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} \).
We know that \( \tan 45^\circ = 1 \).
Now, substitute this value into the expression:
\( \frac{1 - (1)^2}{1 + (1)^2} \)
\( = \frac{1 - 1}{1 + 1} \)
\( = \frac{0}{2} \)
\( = 0 \). This is a straightforward application of a known trigonometric value.
In simple words: Since \( \tan 45^\circ \) is 1, we replace \( \tan^2 45^\circ \) with \( 1^2 \), which is also 1. Then the top part becomes \( 1-1=0 \), and the bottom part becomes \( 1+1=2 \). So, the answer is \( \frac{0}{2} \), which is 0.

🎯 Exam Tip: Always remember that \( \tan 45^\circ = 1 \). This value appears frequently in trigonometry and simplifies calculations quickly.

Question 8. The value of \( \cosec (70^\circ + \theta) - \sec (20^\circ - \theta) + \tan (65^\circ + \theta) - \cot (25^\circ - \theta) \) is .........
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a) 0
We need to simplify the expression \( \cosec (70^\circ + \theta) - \sec (20^\circ - \theta) + \tan (65^\circ + \theta) - \cot (25^\circ - \theta) \).
Let's use the complementary angle identities:
1. \( \sec A = \cosec (90^\circ - A) \)
So, \( \sec (20^\circ - \theta) = \cosec [90^\circ - (20^\circ - \theta)] \)
\( = \cosec (90^\circ - 20^\circ + \theta) \)
\( = \cosec (70^\circ + \theta) \).
2. \( \cot A = \tan (90^\circ - A) \)
So, \( \cot (25^\circ - \theta) = \tan [90^\circ - (25^\circ - \theta)] \)
\( = \tan (90^\circ - 25^\circ + \theta) \)
\( = \tan (65^\circ + \theta) \).
Now, substitute these simplified terms back into the original expression:
\( \cosec (70^\circ + \theta) - \cosec (70^\circ + \theta) + \tan (65^\circ + \theta) - \tan (65^\circ + \theta) \)
All terms cancel out.
\( = 0 \). This is a classic example of using angle relationships to simplify complex trigonometric expressions.
In simple words: We change \( \sec (20^\circ - \theta) \) to \( \cosec (70^\circ + \theta) \) and \( \cot (25^\circ - \theta) \) to \( \tan (65^\circ + \theta) \). After these changes, the positive and negative parts of the expression become exactly the same, making the whole thing equal to zero.

🎯 Exam Tip: For problems with angles involving \( \theta \) or other variables, check if the angles are complementary. For example, \( (70^\circ + \theta) + (20^\circ - \theta) = 90^\circ \), which means they are complementary.

Question 9. The value of \( \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \dots \tan 89^\circ \) is .........
(a) 0
(b) 1
(d) \( \frac{\sqrt{3}}{2} \)
Answer: (b) 1
We need to find the value of the product \( \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \dots \tan 89^\circ \).
We know that \( \tan \theta = \cot (90^\circ - \theta) \).
Let's pair up the terms in the product:
\( \tan 1^\circ \) with \( \tan 89^\circ \)
\( \tan 2^\circ \) with \( \tan 88^\circ \)
...
\( \tan 44^\circ \) with \( \tan 46^\circ \)
The middle term will be \( \tan 45^\circ \).
Using the identity \( \tan \theta = \cot (90^\circ - \theta) \):
\( \tan 89^\circ = \cot (90^\circ - 89^\circ) = \cot 1^\circ \)
\( \tan 88^\circ = \cot (90^\circ - 88^\circ) = \cot 2^\circ \)
...
\( \tan 46^\circ = \cot (90^\circ - 46^\circ) = \cot 44^\circ \)
Now, rewrite the product:
\( (\tan 1^\circ \cdot \cot 1^\circ) \cdot (\tan 2^\circ \cdot \cot 2^\circ) \dots (\tan 44^\circ \cdot \cot 44^\circ) \cdot \tan 45^\circ \)
Since \( \tan \theta \cdot \cot \theta = 1 \), each paired term is equal to 1.
\( = (1) \cdot (1) \dots (1) \cdot \tan 45^\circ \)
We know that \( \tan 45^\circ = 1 \).
So, the entire product is \( 1 \). This type of problem often appears in exams to test understanding of complementary angles and trigonometric identities.
In simple words: When you multiply tangents of angles that add up to \( 90^\circ \), they cancel each other out to 1. For example, \( \tan 1^\circ \) times \( \tan 89^\circ \) is 1. All such pairs become 1, and the middle term \( \tan 45^\circ \) is also 1, so the whole product is 1.

🎯 Exam Tip: In products of trigonometric ratios where angles are in increasing order, always look for complementary pairs. Terms like \( \tan \theta \cdot \tan (90^\circ - \theta) \) simplify to 1, and \( \tan 45^\circ \) is often the unpaired middle term.

Question 10. Given that \( \sin \alpha = \frac{1}{2} \) and \( \cos \beta = \frac{1}{2} \), then the value of \( \alpha + \beta \) is .........
(a) \( 0^\circ \)
(b) \( 90^\circ \)
(c) \( 30^\circ \)
(d) \( 60^\circ \)
Answer: (b) \( 90^\circ \)
Given \( \sin \alpha = \frac{1}{2} \).
We know that \( \sin 30^\circ = \frac{1}{2} \).
Therefore, \( \alpha = 30^\circ \).
Given \( \cos \beta = \frac{1}{2} \).
We know that \( \cos 60^\circ = \frac{1}{2} \).
Therefore, \( \beta = 60^\circ \).
Now, we need to find the value of \( \alpha + \beta \).
\( \alpha + \beta = 30^\circ + 60^\circ \)
\( = 90^\circ \). This problem relies on recognizing standard trigonometric values and basic angle addition.
In simple words: We find the angle \( \alpha \) that has a sine of \( \frac{1}{2} \), which is \( 30^\circ \). Then we find the angle \( \beta \) that has a cosine of \( \frac{1}{2} \), which is \( 60^\circ \). Adding these two angles gives \( 30^\circ + 60^\circ = 90^\circ \).

🎯 Exam Tip: Memorize the sine and cosine values for common angles like 0°, 30°, 45°, 60°, and 90°. These basic values are fundamental for solving many trigonometry problems quickly and accurately.