Samacheer Kalvi Class 9 Maths Solutions Chapter 6 Trigonometry Exercise 6.2

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1. Verify the following equalities:
(i) \( \sin^2 60^\circ + \cos^2 60^\circ = 1 \)
(ii) \( 1 + \tan^2 30^\circ = \sec^2 30^\circ \)
(iii) \( \cos 90^\circ = 1 – 2\sin^2 45^\circ = 2\cos^2 45^\circ – 1 \)
(iv) \( \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ = \sin 90^\circ \)
Answer:
(i) To verify \( \sin^2 60^\circ + \cos^2 60^\circ = 1 \):
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 60^\circ = \frac{1}{2} \).
Let's take the Left Hand Side (L.H.S.):
L.H.S \( = \sin^2 60^\circ + \cos^2 60^\circ \)
\( = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 \)
\( = \frac{3}{4} + \frac{1}{4} \)
\( = \frac{3+1}{4} \)
\( = \frac{4}{4} \)
\( = 1 \)
Since L.H.S \( = 1 \) and R.H.S \( = 1 \), we can say L.H.S \( = \) R.H.S.
Thus, the equality is proved.

(ii) To verify \( 1 + \tan^2 30^\circ = \sec^2 30^\circ \):
We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \sec 30^\circ = \frac{2}{\sqrt{3}} \).
Let's take the Left Hand Side (L.H.S.):
L.H.S \( = 1 + \tan^2 30^\circ \)
\( = 1 + \left(\frac{1}{\sqrt{3}}\right)^2 \)
\( = 1 + \frac{1}{3} \)
\( = \frac{3+1}{3} \)
\( = \frac{4}{3} \)
Now, let's take the Right Hand Side (R.H.S.):
R.H.S \( = \sec^2 30^\circ \)
\( = \left(\frac{2}{\sqrt{3}}\right)^2 \)
\( = \frac{4}{3} \)
Since L.H.S \( = \frac{4}{3} \) and R.H.S \( = \frac{4}{3} \), we can say L.H.S \( = \) R.H.S.
Thus, the equality is proved.

(iii) To verify \( \cos 90^\circ = 1 – 2\sin^2 45^\circ = 2\cos^2 45^\circ – 1 \):
We know that \( \cos 90^\circ = 0 \), \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), and \( \cos 45^\circ = \frac{1}{\sqrt{2}} \).
Let's check the first part:
\( \cos 90^\circ = 0 \) ...(1)
Now, let's check the second part:
\( 1 – 2\sin^2 45^\circ = 1 - 2\left(\frac{1}{\sqrt{2}}\right)^2 \)
\( = 1 - 2 \times \frac{1}{2} \)
\( = 1 - 1 \)
\( = 0 \) ...(2)
Finally, let's check the third part:
\( 2\cos^2 45^\circ – 1 = 2\left(\frac{1}{\sqrt{2}}\right)^2 - 1 \)
\( = 2 \times \frac{1}{2} - 1 \)
\( = 1 - 1 \)
\( = 0 \) ...(3)
From equations (1), (2), and (3), we see that all three parts are equal to 0.
Therefore, \( \cos 90^\circ = 1 – 2\sin^2 45^\circ = 2\cos^2 45^\circ – 1 \).
Thus, the equality is proved.

(iv) To verify \( \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ = \sin 90^\circ \):
We know the values:
\( \sin 30^\circ = \frac{1}{2} \)
\( \cos 60^\circ = \frac{1}{2} \)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 90^\circ = 1 \)
Let's take the Left Hand Side (L.H.S.):
L.H.S \( = \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ \)
\( = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{1}{4} + \frac{3}{4} \)
\( = \frac{1+3}{4} \)
\( = \frac{4}{4} \)
\( = 1 \)
Now, let's take the Right Hand Side (R.H.S.):
R.H.S \( = \sin 90^\circ \)
\( = 1 \)
Since L.H.S \( = 1 \) and R.H.S \( = 1 \), we can say L.H.S \( = \) R.H.S.
Thus, the equality is proved.
In simple words: For each part, we replace the trigonometry terms with their known fraction values. Then, we do the math on both sides of the equals sign. If both sides give the same number, the equality is correct. This shows that the trigonometric identities hold true for the given angles.

🎯 Exam Tip: When verifying trigonometric equalities, always list the exact values of the trigonometric ratios for the given angles first. This helps avoid calculation errors and makes your solution clear.

Question 2. Find the value of the following:
(i) \( \frac{\tan 45^\circ}{\operatorname{cosec} 30^\circ} + \frac{\sec 60^\circ}{\cot 45^\circ} – \frac{5 \sin 90^\circ}{2 \cos 0^\circ} \)
(ii) \( (\sin 90^\circ + \cos 60^\circ + \cos 45^\circ) \times (\sin 30^\circ + \cos 0^\circ - \cos 45^\circ) \)
(iii) \( \sin^2 30^\circ – 2 \cos^3 60^\circ + 3 \tan^4 45^\circ \)
Answer:
We list the common trigonometric values needed for these problems:
\( \sin 30^\circ = \frac{1}{2} \)
\( \cos 60^\circ = \frac{1}{2} \)
\( \tan 45^\circ = 1 \)
\( \operatorname{cosec} 30^\circ = 2 \)
\( \sec 60^\circ = 2 \)
\( \cot 45^\circ = 1 \)
\( \sin 90^\circ = 1 \)
\( \cos 0^\circ = 1 \)
\( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

(i) To find the value of \( \frac{\tan 45^\circ}{\operatorname{cosec} 30^\circ} + \frac{\sec 60^\circ}{\cot 45^\circ} – \frac{5 \sin 90^\circ}{2 \cos 0^\circ} \):
Substitute the known values into the expression:
\( = \frac{1}{2} + \frac{2}{1} - \frac{5(1)}{2(1)} \)
\( = \frac{1}{2} + 2 - \frac{5}{2} \)
Combine the terms with a common denominator:
\( = \frac{1+4-5}{2} \)
\( = \frac{0}{2} \)
\( = 0 \)

(ii) To find the value of \( (\sin 90^\circ + \cos 60^\circ + \cos 45^\circ) \times (\sin 30^\circ + \cos 0^\circ - \cos 45^\circ) \):
Substitute the known values into the expression:
\( = \left(1 + \frac{1}{2} + \frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{2} + 1 - \frac{1}{\sqrt{2}}\right) \)
First, simplify each bracket:
For the first bracket: \( \left(1 + \frac{1}{2} + \frac{1}{\sqrt{2}}\right) = \left(\frac{2+1}{2} + \frac{1}{\sqrt{2}}\right) = \left(\frac{3}{2} + \frac{1}{\sqrt{2}}\right) = \left(\frac{3\sqrt{2}}{2\sqrt{2}} + \frac{2}{2\sqrt{2}}\right) = \frac{3\sqrt{2}+2}{2\sqrt{2}} \)
For the second bracket: \( \left(\frac{1}{2} + 1 - \frac{1}{\sqrt{2}}\right) = \left(\frac{1+2}{2} - \frac{1}{\sqrt{2}}\right) = \left(\frac{3}{2} - \frac{1}{\sqrt{2}}\right) = \left(\frac{3\sqrt{2}}{2\sqrt{2}} - \frac{2}{2\sqrt{2}}\right) = \frac{3\sqrt{2}-2}{2\sqrt{2}} \)
Now, multiply the simplified brackets:
\( = \left(\frac{3\sqrt{2}+2}{2\sqrt{2}}\right) \times \left(\frac{3\sqrt{2}-2}{2\sqrt{2}}\right) \)
Use the identity \( (a+b)(a-b) = a^2 - b^2 \) for the numerator:
\( = \frac{(3\sqrt{2})^2 - (2)^2}{(2\sqrt{2})^2} \)
\( = \frac{(9 \times 2) - 4}{(4 \times 2)} \)
\( = \frac{18 - 4}{8} \)
\( = \frac{14}{8} \)
\( = \frac{7}{4} \)

(iii) To find the value of \( \sin^2 30^\circ – 2 \cos^3 60^\circ + 3 \tan^4 45^\circ \):
Substitute the known values into the expression:
\( = \left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^3 + 3(1)^4 \)
Calculate the powers:
\( = \frac{1}{4} - 2\left(\frac{1}{8}\right) + 3(1) \)
Simplify the terms:
\( = \frac{1}{4} - \frac{2}{8} + 3 \)
\( = \frac{1}{4} - \frac{1}{4} + 3 \)
\( = 0 + 3 \)
\( = 3 \)
In simple words: For each problem, we put the correct number for each trigonometric function (like sin 30 degrees) into the equation. Then we follow the order of operations (like squaring or cubing first, then multiplying, then adding or subtracting) to find the final answer. This involves careful calculation of fractions and roots.

🎯 Exam Tip: Always write down all the necessary trigonometric ratios at the start. This prevents errors when substituting values into complex expressions and helps organize your work.

Question 3. Verify \( \cos 3 A = 4 \cos^3 A – 3 \cos A \), when \( A = 30^\circ \)
Answer:
We need to verify the identity \( \cos 3 A = 4 \cos^3 A – 3 \cos A \) by setting \( A = 30^\circ \).
First, let's calculate the Left Hand Side (L.H.S.):
L.H.S \( = \cos 3 A \)
Substitute \( A = 30^\circ \):
\( = \cos (3 \times 30^\circ) \)
\( = \cos 90^\circ \)
We know that \( \cos 90^\circ = 0 \).
So, L.H.S \( = 0 \).

Next, let's calculate the Right Hand Side (R.H.S.):
R.H.S \( = 4 \cos^3 A – 3 \cos A \)
Substitute \( A = 30^\circ \):
\( = 4 \cos^3 30^\circ – 3 \cos 30^\circ \)
We know that \( \cos 30^\circ = \frac{\sqrt{3}}{2} \).
\( = 4 \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \left(\frac{\sqrt{3}}{2}\right) \)
\( = 4 \left(\frac{3\sqrt{3}}{8}\right) - \frac{3\sqrt{3}}{2} \)
\( = \frac{4 \times 3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} \)
\( = \frac{12\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} \)
Simplify the first term:
\( = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} \)
\( = 0 \)
So, R.H.S \( = 0 \).
Since L.H.S \( = 0 \) and R.H.S \( = 0 \), we can say L.H.S \( = \) R.H.S.
Therefore, the equality \( \cos 3 A = 4 \cos^3 A – 3 \cos A \) is verified for \( A = 30^\circ \). The formula for the cosine of a triple angle is correctly demonstrated.
In simple words: We check if a special trigonometry rule works. We put the number 30 degrees into both sides of the rule where 'A' is. Then we do the math for each side. If both sides end up with the same answer (which is 0 in this case), then the rule is proven correct for that angle.

🎯 Exam Tip: When verifying identities, always calculate the Left Hand Side and Right Hand Side separately. This systematic approach helps prevent errors and ensures all steps are clearly shown.

Question 4. Find the value of \( 8 \sin 2x \cos 4x \sin 6x \), when \( x = 15^\circ \).
Answer:
We need to find the value of the expression \( 8 \sin 2x \cos 4x \sin 6x \) when \( x = 15^\circ \).
First, substitute \( x = 15^\circ \) into the expression:
\( = 8 \sin (2 \times 15^\circ) \cos (4 \times 15^\circ) \sin (6 \times 15^\circ) \)
Calculate the angles:
\( = 8 \sin 30^\circ \cos 60^\circ \sin 90^\circ \)
Now, substitute the known trigonometric values:
We know that \( \sin 30^\circ = \frac{1}{2} \), \( \cos 60^\circ = \frac{1}{2} \), and \( \sin 90^\circ = 1 \).
\( = 8 \times \frac{1}{2} \times \frac{1}{2} \times 1 \)
Multiply the numbers:
\( = 8 \times \frac{1}{4} \times 1 \)
\( = 2 \)
Therefore, the value of the expression is 2. This shows how knowing basic trigonometric values helps simplify complex expressions quickly.
In simple words: We are given a math problem with 'x' in it. First, we put 15 degrees in place of 'x'. Then, we find the values of sin, cos, and sin for the new angles. Finally, we multiply all the numbers together to get the final answer.

🎯 Exam Tip: Always simplify the angles inside the trigonometric functions before looking up their values. This ensures you use the correct standard angle ratios.