Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.1

Question 1. Identify the quadrant in which an angle of each given measure lies,
(i) 25°
(ii) 825°
(iii) – 55°
(iv) 328°
(v) – 230°
Answer:
(i) 25°The angle 25° lies in the first quadrant. This is because it is a positive angle between 0° and 90°. When angles are measured from the positive x-axis in a counter-clockwise direction, the first 90 degrees fall into the first quadrant.
In simple words: The angle 25° is in the first quarter (quadrant) because it is a positive angle between 0 and 90 degrees.
(ii) 825°To find the quadrant for 825°, we can see how many full rotations of 360° it contains. \( 825^\circ = 2 \times 360^\circ + 105^\circ \). After two full rotations, the remaining angle is 105°. Since 105° is between 90° and 180°, the angle 825° lies in the second quadrant. Each full rotation brings the angle back to the starting position, making it easier to identify the quadrant.
In simple words: First, take out full circles of 360 degrees from 825. You are left with 105 degrees. This 105 degree angle sits in the second quarter of the circle.
(iii) – 55°The angle \( -55^\circ \) is measured clockwise from the positive x-axis. A negative angle between \( 0^\circ \) and \( -90^\circ \) falls into the fourth quadrant. So, \( -55^\circ \) is in the fourth quadrant. Negative angles rotate in the opposite direction from positive angles, which rotate counter-clockwise.
In simple words: For -55 degrees, you measure clockwise. This angle falls in the fourth quarter of the circle.
(iv) 328°The angle \( 328^\circ \) is a positive angle. Since it is greater than 270° but less than 360°, it is located in the fourth quadrant. We can also write it as \( 270^\circ + 58^\circ \). This means it has passed 270 degrees and gone another 58 degrees, which puts it into the last section of the circle.
In simple words: The 328-degree angle is bigger than 270 but less than 360. This places it in the fourth quarter.
(v) – 230°For \( -230^\circ \), we measure clockwise. \( -230^\circ \) is equivalent to \( -180^\circ \) plus another \( -50^\circ \). This means it goes through the third quadrant completely (from 0 to -180) and then enters the second quadrant (from -180 to -270). So, \( -230^\circ \) lies in the second quadrant. Understanding the direction of rotation is key for negative angles.
In simple words: For -230 degrees, rotate clockwise. You pass -180 degrees and go another 50 degrees, landing in the second quarter.

🎯 Exam Tip: To identify the quadrant for large positive or negative angles, add or subtract multiples of 360° until the angle falls within the 0° to 360° range.

Question 2. For each given angle, find a co-terminal angle with a measure of \( \theta \) such that \( 0 \le \theta < 360^\circ \).
(i) 395°
(ii) 525°
(iii) 1150°
(iv) – 270°
(v) – 450°
Answer:
(i) 395°
To find a coterminal angle between 0° and 360°, we subtract multiples of 360°. For \( 395^\circ \), we have \( 395^\circ = 360^\circ + 35^\circ \). So, the coterminal angle is \( 35^\circ \). Coterminal angles share the same terminal side when drawn in standard position.
In simple words: Subtract 360 from 395. The angle left is 35 degrees. This is the coterminal angle.
(ii) 525°
For \( 525^\circ \), subtract 360°: \( 525^\circ = 360^\circ + 165^\circ \). The coterminal angle in the desired range is \( 165^\circ \). This means 525 degrees has gone one full circle and then an additional 165 degrees.
In simple words: Take away 360 from 525. The leftover angle, 165 degrees, is the coterminal angle.
(iii) 1150°
For \( 1150^\circ \), we can subtract 360° multiple times. \( 1150^\circ = 3 \times 360^\circ + 70^\circ \). The coterminal angle is \( 70^\circ \). This shows that 1150 degrees makes three complete turns and then stops at 70 degrees.
In simple words: Subtract 360 degrees three times from 1150. What is left, 70 degrees, is the coterminal angle.
(iv) – 270°
For \( -270^\circ \), we add multiples of 360° until the angle is positive and within the range. \( -270^\circ + 360^\circ = 90^\circ \). So, the coterminal angle is \( 90^\circ \). Adding 360 degrees to a negative angle helps find its positive equivalent.
In simple words: Add 360 to -270. The angle 90 degrees is the coterminal angle.
(v) – 450°
For \( -450^\circ \), we add 360° multiple times. \( -450^\circ + 2 \times 360^\circ = -450^\circ + 720^\circ = 270^\circ \). The coterminal angle is \( 270^\circ \). You need to add enough full rotations to make the angle positive and within the desired range.
In simple words: Add 360 twice to -450. You get 270 degrees, which is the coterminal angle.

🎯 Exam Tip: For positive angles, subtract multiples of 360° until the result is between 0° and 360°. For negative angles, add multiples of 360° until the result is in the same range.

Question 3. If \( a \cos \theta - b \sin \theta = c \), show that \( a \sin \theta + b \cos \theta = \pm \sqrt{a^2+b^2-c^2} \)
Answer:
We are given the equation \( a \cos \theta - b \sin \theta = c \) ...(1). We need to show that \( a \sin \theta + b \cos \theta = \pm \sqrt{a^2+b^2-c^2} \). Let's call the second expression \( x \), so \( a \sin \theta + b \cos \theta = x \) ...(2).
Now, we square both the given equation and our new expression:
\( (a \cos \theta - b \sin \theta)^2 = c^2 \)
\( \implies a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta = c^2 \) ...(3)
\( (a \sin \theta + b \cos \theta)^2 = x^2 \)
\( \implies a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = x^2 \) ...(4)
When we add these two expanded equations (3) and (4), the terms \( -2ab \sin \theta \cos \theta \) and \( +2ab \sin \theta \cos \theta \) cancel each other out.
Adding (3) and (4):
\( (a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta) = c^2 + x^2 \)
\( \implies a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = c^2 + x^2 \)
Since we know that \( \cos^2 \theta + \sin^2 \theta = 1 \), this simplifies to:
\( a^2 (1) + b^2 (1) = c^2 + x^2 \)
\( \implies a^2 + b^2 = c^2 + x^2 \)
From this, we can find \( x^2 \):
\( x^2 = a^2 + b^2 - c^2 \)
Taking the square root of both sides, we get:
\( x = \pm \sqrt{a^2 + b^2 - c^2} \)
By substituting \( x \) back, we prove the identity: \( a \sin \theta + b \cos \theta = \pm \sqrt{a^2+b^2-c^2} \). This method uses a fundamental trigonometric identity to simplify the problem, showing how squaring and adding can eliminate certain terms.
In simple words: We are given one math rule with 'c' and want to prove another rule for 'x'. If we square both rules and add them together, some parts with \( \sin \theta \cos \theta \) will cancel out. We use the basic rule that \( \sin^2 \theta + \cos^2 \theta \) equals 1. After doing all the math, we find that 'x' is equal to plus or minus the square root of \( a^2 + b^2 - c^2 \).

🎯 Exam Tip: When asked to prove relationships involving sums and differences of trigonometric terms, squaring both sides and using \( \sin^2 \theta + \cos^2 \theta = 1 \) is often a useful strategy.

Question 4. If \( \sin \theta + \cos \theta = m \), show that \( \cos^6 \theta + \sin^6 \theta = \frac{4-3(m^2-1)^2}{4} \) where \( m^2 \le 2 \).
Answer:
We are given \( \sin \theta + \cos \theta = m \). We need to prove the identity for \( \cos^6 \theta + \sin^6 \theta \).
First, we square the given equation:
\( (\sin \theta + \cos \theta)^2 = m^2 \)
\( \implies \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = m^2 \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we get:
\( 1 + 2 \sin \theta \cos \theta = m^2 \)
From this, we can find \( \sin \theta \cos \theta \):
\( 2 \sin \theta \cos \theta = m^2 - 1 \)
\( \implies \sin \theta \cos \theta = \frac{m^2 - 1}{2} \) ...(1)
Next, we rewrite \( \cos^6 \theta + \sin^6 \theta \) using the algebraic identity \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \).
Here, let \( a = \cos^2 \theta \) and \( b = \sin^2 \theta \).
\( \cos^6 \theta + \sin^6 \theta = (\cos^2 \theta)^3 + (\sin^2 \theta)^3 \)
\( \implies = (\cos^2 \theta + \sin^2 \theta) [(\cos^2 \theta)^2 - \cos^2 \theta \sin^2 \theta + (\sin^2 \theta)^2] \)
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), this simplifies to:
\( = (1) [(\cos^2 \theta)^2 + (\sin^2 \theta)^2 - \cos^2 \theta \sin^2 \theta] \)
Now, we use another algebraic identity: \( a^2 + b^2 = (a+b)^2 - 2ab \).
Applying this to \( (\cos^2 \theta)^2 + (\sin^2 \theta)^2 \):
\( (\cos^2 \theta)^2 + (\sin^2 \theta)^2 = (\cos^2 \theta + \sin^2 \theta)^2 - 2 \cos^2 \theta \sin^2 \theta \)
\( \implies = (1)^2 - 2 \cos^2 \theta \sin^2 \theta \)
\( \implies = 1 - 2 \cos^2 \theta \sin^2 \theta \)
Substitute this back into the expression for \( \cos^6 \theta + \sin^6 \theta \):
\( \cos^6 \theta + \sin^6 \theta = (1 - 2 \cos^2 \theta \sin^2 \theta) - \cos^2 \theta \sin^2 \theta \)
\( \implies = 1 - 3 \cos^2 \theta \sin^2 \theta \)
Now, we substitute the value of \( \sin \theta \cos \theta \) we found in (1) into this expression:
\( \implies = 1 - 3 \left( \frac{m^2 - 1}{2} \right)^2 \)
\( \implies = 1 - 3 \frac{(m^2 - 1)^2}{4} \)
To combine these, find a common denominator:
\( \implies = \frac{4}{4} - \frac{3(m^2 - 1)^2}{4} \)
\( \implies = \frac{4 - 3(m^2 - 1)^2}{4} \)
This matches the right-hand side, thus proving the identity. This proof involves using both algebraic identities for cubes and squares, along with fundamental trigonometric identities.
In simple words: First, we start with the given rule and square both sides to find a value for \( \sin \theta \cos \theta \). Then, we look at the part we need to prove, \( \cos^6 \theta + \sin^6 \theta \). We can rewrite this as \( (\cos^2 \theta)^3 + (\sin^2 \theta)^3 \). We use special algebra rules to break this down. Once it's simplified, we put back the value of \( \sin \theta \cos \theta \) we found at the beginning. After some more simple steps, we will get the answer we need to prove.

🎯 Exam Tip: Remember common algebraic identities like \( a^3+b^3 \) and \( a^2+b^2 \), as they frequently appear in trigonometric proofs alongside identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 5. If \( \frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1 \), Prove that
(i) \( \sin^4 \alpha + \sin^4 \beta = 2 \sin^2 \alpha \sin^2 \beta \)
(ii) \( \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} = 1 \)
Answer:
We begin with the given equation: \( \frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1 \).
(i) To prove \( \sin^4 \alpha + \sin^4 \beta = 2 \sin^2 \alpha \sin^2 \beta \):
We convert \( \cos \) terms into \( \sin \) terms using the identity \( \cos^2 x = 1 - \sin^2 x \).
So, \( \cos^4 \alpha = (1-\sin^2 \alpha)^2 \) and \( \cos^2 \beta = 1-\sin^2 \beta \).
Substitute these into the given equation:
\( \frac{(1-\sin^2 \alpha)^2}{1-\sin^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1 \)
Now, we find a common denominator, which is \( (1-\sin^2 \beta)\sin^2 \beta \). Multiply both sides by this common denominator to clear the fractions:
\( \sin^2 \beta (1-\sin^2 \alpha)^2 + \sin^4 \alpha (1-\sin^2 \beta) = (1-\sin^2 \beta)\sin^2 \beta \)
Expand the terms, especially \( (1-\sin^2 \alpha)^2 = 1 - 2\sin^2 \alpha + \sin^4 \alpha \):
\( \sin^2 \beta (1 - 2\sin^2 \alpha + \sin^4 \alpha) + \sin^4 \alpha - \sin^4 \alpha \sin^2 \beta = \sin^2 \beta - \sin^4 \beta \)
\( \implies \sin^2 \beta - 2\sin^2 \alpha \sin^2 \beta + \sin^4 \alpha \sin^2 \beta + \sin^4 \alpha - \sin^4 \alpha \sin^2 \beta = \sin^2 \beta - \sin^4 \beta \)
The terms \( \sin^4 \alpha \sin^2 \beta \) cancel out:
\( \implies \sin^2 \beta - 2\sin^2 \alpha \sin^2 \beta + \sin^4 \alpha = \sin^2 \beta - \sin^4 \beta \)
Cancel \( \sin^2 \beta \) from both sides:
\( \implies -2\sin^2 \alpha \sin^2 \beta + \sin^4 \alpha = - \sin^4 \beta \)
Rearranging the terms, we get:
\( \sin^4 \alpha + \sin^4 \beta = 2 \sin^2 \alpha \sin^2 \beta \)
This proves the first part. This transformation shows how one trigonometric identity can be transformed into another through careful algebraic manipulation.
In simple words: Start with the given math rule. Change all the 'cos' parts into 'sin' parts using the rule \( \cos^2 x = 1 - \sin^2 x \). Then, clear the fractions and multiply everything out. Many parts will cancel each other. After moving terms around, you will see that \( \sin^4 \alpha + \sin^4 \beta \) is equal to \( 2 \sin^2 \alpha \sin^2 \beta \).
(ii) To prove \( \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} = 1 \):
From the result of part (i), we have \( \sin^4 \alpha + \sin^4 \beta = 2 \sin^2 \alpha \sin^2 \beta \).
We can rearrange this equation:
\( \sin^4 \alpha - 2 \sin^2 \alpha \sin^2 \beta + \sin^4 \beta = 0 \)
This is in the form of \( (a-b)^2 = 0 \), where \( a = \sin^2 \alpha \) and \( b = \sin^2 \beta \).
\( \implies (\sin^2 \alpha - \sin^2 \beta)^2 = 0 \)
This implies that \( \sin^2 \alpha - \sin^2 \beta = 0 \), so \( \sin^2 \alpha = \sin^2 \beta \).
Because of the identity \( \sin^2 x + \cos^2 x = 1 \), if \( \sin^2 \alpha = \sin^2 \beta \), then it logically follows that \( 1 - \cos^2 \alpha = 1 - \cos^2 \beta \), which means \( \cos^2 \alpha = \cos^2 \beta \).
Now, we take the expression we need to prove:
\( \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} \)
We substitute \( \cos^2 \beta \) with \( \cos^2 \alpha \) and \( \sin^2 \beta \) with \( \sin^2 \alpha \) (since they are equal from our derived relation):
\( \implies = \frac{(\cos^2 \alpha)^2}{\cos^2 \alpha} + \frac{(\sin^2 \alpha)^2}{\sin^2 \alpha} \)
This simplifies to:
\( \implies = \cos^2 \alpha + \sin^2 \alpha \)
Finally, using the fundamental identity, \( \cos^2 \alpha + \sin^2 \alpha = 1 \).
Thus, \( \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} = 1 \). The second part is also proven. This demonstrates how derived relationships can be used to prove further identities efficiently.
In simple words: From the first part, we found that \( \sin^2 \alpha = \sin^2 \beta \). This also means \( \cos^2 \alpha = \cos^2 \beta \). Now, we take the second thing we need to prove. We replace the 'beta' parts with 'alpha' parts because they are equal. The math then becomes very simple: \( \cos^2 \alpha + \sin^2 \alpha \), which always equals 1. So, the second part is also shown to be true.

🎯 Exam Tip: When a problem has multiple parts, the result of an earlier part often simplifies the solution for subsequent parts. Always look for connections.

Question 6. If \( y = \frac{2 \sin \alpha}{1 + \cos \alpha + \sin \alpha} \), Prove that \( y = \frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha} \)
Answer:
We are given the expression for \( y \) as \( y = \frac{2 \sin \alpha}{1 + \cos \alpha + \sin \alpha} \). To prove the desired identity, we use a common algebraic technique: multiplying the numerator and denominator by the conjugate of the denominator.
The denominator can be thought of as \( (1 + \sin \alpha) + \cos \alpha \). Its conjugate is \( (1 + \sin \alpha) - \cos \alpha \).
Multiply both the numerator and denominator by this term:
\( y = \frac{2 \sin \alpha}{ (1 + \sin \alpha) + \cos \alpha } \times \frac{ (1 + \sin \alpha) - \cos \alpha }{ (1 + \sin \alpha) - \cos \alpha } \)
Let's simplify the denominator first. It is of the form \( (A+B)(A-B) = A^2 - B^2 \), where \( A = (1 + \sin \alpha) \) and \( B = \cos \alpha \).
Denominator \( = (1 + \sin \alpha)^2 - \cos^2 \alpha \)
Expand \( (1 + \sin \alpha)^2 \):
\( = (1 + 2 \sin \alpha + \sin^2 \alpha) - \cos^2 \alpha \)
Now, replace \( \cos^2 \alpha \) with \( 1 - \sin^2 \alpha \):
\( = 1 + 2 \sin \alpha + \sin^2 \alpha - (1 - \sin^2 \alpha) \)
\( = 1 + 2 \sin \alpha + \sin^2 \alpha - 1 + \sin^2 \alpha \)
Combine like terms:
\( = 2 \sin \alpha + 2 \sin^2 \alpha \)
Factor out \( 2 \sin \alpha \):
\( = 2 \sin \alpha (1 + \sin \alpha) \)
Now, let's look at the numerator:
Numerator \( = 2 \sin \alpha ((1 + \sin \alpha) - \cos \alpha) \)
So, the expression for \( y \) becomes:
\( y = \frac{2 \sin \alpha (1 - \cos \alpha + \sin \alpha)}{2 \sin \alpha (1 + \sin \alpha)} \)
We can cancel out the common term \( 2 \sin \alpha \) from both the numerator and denominator (assuming \( \sin \alpha \neq 0 \)):
\( y = \frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha} \)
This is the required identity. This method elegantly transforms complex expressions by rationalizing the denominator, making it simpler to prove the equivalence.
In simple words: We start with the given value of \( y \). To change it into the form we want, we multiply the top and bottom by a special term that helps us simplify. This term is like the bottom part but with a minus sign in the middle. The bottom part then turns into \( 2 \sin \alpha (1 + \sin \alpha) \) after some math steps and using the rule \( \sin^2 x + \cos^2 x = 1 \). The top part also gets multiplied. Then, we can cross out the same part, \( 2 \sin \alpha \), from both the top and bottom. This leaves us with the answer we needed to prove.

🎯 Exam Tip: When the denominator of a fraction contains a sum or difference, multiplying by the conjugate is a powerful technique for simplification, especially when trigonometric terms are involved.

Question 7. If \( x = \sum_{n=0}^{\infty} \cos^{2n} \theta \), \( y = \sum_{n=0}^{\infty} \sin^{2n} \theta \) and \( z = \sum_{n=0}^{\infty} \cos^{2n} \theta \sin^{2n} \theta \), \( 0 < \theta < \frac{\pi}{2} \), then show that \( xyz = x + y + z \).
Answer:
We are given three infinite series, \( x \), \( y \), and \( z \), and need to prove that their product equals their sum. Each series is a geometric progression. The sum of an infinite geometric series is given by the formula \( S = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio. Since \( 0 < \theta < \frac{\pi}{2} \), we know that \( 0 < \sin \theta < 1 \) and \( 0 < \cos \theta < 1 \), which ensures that \( |r| < 1 \) for each series.
For \( x \):
\( x = 1 + \cos^2 \theta + \cos^4 \theta + \ldots \)
Here, the first term \( a = 1 \) and the common ratio \( r = \cos^2 \theta \).
\( x = \frac{1}{1 - \cos^2 \theta} \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have \( 1 - \cos^2 \theta = \sin^2 \theta \).
\( \implies x = \frac{1}{\sin^2 \theta} \) ...(1)
For \( y \):
\( y = 1 + \sin^2 \theta + \sin^4 \theta + \ldots \)
Here, the first term \( a = 1 \) and the common ratio \( r = \sin^2 \theta \).
\( y = \frac{1}{1 - \sin^2 \theta} \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have \( 1 - \sin^2 \theta = \cos^2 \theta \).
\( \implies y = \frac{1}{\cos^2 \theta} \) ...(2)
For \( z \):
\( z = 1 + \cos^2 \theta \sin^2 \theta + \cos^4 \theta \sin^4 \theta + \ldots \)
Here, the first term \( a = 1 \) and the common ratio \( r = \cos^2 \theta \sin^2 \theta \).
\( z = \frac{1}{1 - \cos^2 \theta \sin^2 \theta} \) ...(3)
Now, let's calculate \( x + y + z \):
\( x + y + z = \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} + \frac{1}{1 - \cos^2 \theta \sin^2 \theta} \)
First, combine the first two terms:
\( \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta \cos^2 \theta} \)
\( \implies = \frac{1}{\sin^2 \theta \cos^2 \theta} \)
So, \( x + y + z = \frac{1}{\sin^2 \theta \cos^2 \theta} + \frac{1}{1 - \cos^2 \theta \sin^2 \theta} \)
To combine these, find a common denominator:
\( \implies = \frac{(1 - \cos^2 \theta \sin^2 \theta) + (\sin^2 \theta \cos^2 \theta)}{\sin^2 \theta \cos^2 \theta (1 - \cos^2 \theta \sin^2 \theta)} \)
The terms \( - \cos^2 \theta \sin^2 \theta \) and \( + \sin^2 \theta \cos^2 \theta \) cancel out in the numerator.
\( \implies x + y + z = \frac{1}{\sin^2 \theta \cos^2 \theta (1 - \cos^2 \theta \sin^2 \theta)} \) ...(4)
Next, let's calculate the product \( xyz \):
\( xyz = \left( \frac{1}{\sin^2 \theta} \right) \left( \frac{1}{\cos^2 \theta} \right) \left( \frac{1}{1 - \cos^2 \theta \sin^2 \theta} \right) \)
\( \implies xyz = \frac{1}{\sin^2 \theta \cos^2 \theta (1 - \cos^2 \theta \sin^2 \theta)} \) ...(5)
By comparing the expressions for \( x+y+z \) (4) and \( xyz \) (5), they are identical.
Therefore, \( xyz = x + y + z \). This problem highlights the properties of geometric series when applied to trigonometric functions, showcasing a neat mathematical relationship.
In simple words: We have three special sums called \( x \), \( y \), and \( z \). Each sum follows a pattern. We find a simple form for each sum using a rule for infinite geometric series. We find \( x = \frac{1}{\sin^2 \theta} \), \( y = \frac{1}{\cos^2 \theta} \), and \( z = \frac{1}{1 - \cos^2 \theta \sin^2 \theta} \). Then, we add \( x \), \( y \), and \( z \) together, combining them step by step. After simplifying, we get a long fraction. Next, we multiply \( x \), \( y \), and \( z \) together. We find that the result of the sum and the result of the product are exactly the same. So, we have shown that \( xyz = x+y+z \).

🎯 Exam Tip: Remember the formula for the sum of an infinite geometric series \( S = \frac{a}{1-r} \). This is crucial for problems involving sums of powers like this one.

Question 8. If \( \tan^2 \theta = 1 - k^2 \), show that \( \sec \theta + \tan^3 \theta \operatorname{cosec} \theta = ( 2 - k^2)^{3/2} \). Also, find the values of k for which this result holds.
Answer:
First part: Show the identity.
We are given that \( \tan^2 \theta = 1 - k^2 \). Our goal is to prove the identity \( \sec \theta + \tan^3 \theta \operatorname{cosec} \theta = ( 2 - k^2)^{3/2} \).
First, we use the trigonometric identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
Substituting the given value for \( \tan^2 \theta \):
\( \sec^2 \theta = 1 + (1 - k^2) \)
\( \implies \sec^2 \theta = 2 - k^2 \)
Taking the square root (assuming \( \sec \theta \) is positive, which is valid for \( 0 < \theta < \pi/2 \) typically implied in these problems):
\( \sec \theta = \sqrt{2 - k^2} \)
Now, let's work with the left-hand side (LHS) of the identity we need to prove:
\( \text{LHS} = \sec \theta + \tan^3 \theta \operatorname{cosec} \theta \)
We rewrite \( \operatorname{cosec} \theta \) as \( \frac{1}{\sin \theta} \) and \( \tan \theta \) as \( \frac{\sin \theta}{\cos \theta} \):
\( \implies = \sec \theta + \left(\frac{\sin \theta}{\cos \theta}\right)^3 \times \frac{1}{\sin \theta} \)
\( \implies = \sec \theta + \frac{\sin^3 \theta}{\cos^3 \theta} \times \frac{1}{\sin \theta} \)
Cancel one \( \sin \theta \) term from numerator and denominator:
\( \implies = \sec \theta + \frac{\sin^2 \theta}{\cos^3 \theta} \)
Now, convert \( \sec \theta \) to \( \frac{1}{\cos \theta} \):
\( \implies = \frac{1}{\cos \theta} + \frac{\sin^2 \theta}{\cos^3 \theta} \)
Combine these fractions with a common denominator of \( \cos^3 \theta \):
\( \implies = \frac{\cos^2 \theta}{\cos^3 \theta} + \frac{\sin^2 \theta}{\cos^3 \theta} \)
\( \implies = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^3 \theta} \)
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( \implies = \frac{1}{\cos^3 \theta} \)
We know that \( \frac{1}{\cos \theta} = \sec \theta \), so the expression is:
\( \implies = (\sec \theta)^3 \)
Finally, substitute the value of \( \sec \theta \) we found earlier, \( \sqrt{2 - k^2} \):
\( \implies = (\sqrt{2 - k^2})^3 \)
\( \implies = (2 - k^2)^{3/2} \)
This matches the right-hand side (RHS), thus proving the identity. The key here is to simplify the complex trigonometric expression into a power of secant.
Second part: Find the values of k for which this result holds.
We need to find the values of \( k \) for which this result is valid. We start with the given condition: \( \tan^2 \theta = 1 - k^2 \).
Since the square of any real number (like \( \tan \theta \)) cannot be negative, \( \tan^2 \theta \) must be greater than or equal to zero. Therefore, \( 1 - k^2 \ge 0 \).
This inequality means that \( k^2 \le 1 \).
If \( k^2 \le 1 \), then \( k \) must be between \( -1 \) and \( 1 \), inclusive. So, \( -1 \le k \le 1 \).
Also, for the term \( (2 - k^2)^{3/2} \) to be a real number, the base \( (2 - k^2) \) must be non-negative. This means \( 2 - k^2 \ge 0 \), or \( k^2 \le 2 \).
Since \( k^2 \le 1 \) is a stricter condition than \( k^2 \le 2 \), the values of \( k \) that satisfy all conditions are \( -1 \le k \le 1 \). This ensures that both sides of the equation are well-defined in the real number system.
In simple words: First part: We are given a rule for \( \tan^2 \theta \). We want to show a longer rule is true. We use simple math rules like \( \sec^2 \theta = 1 + \tan^2 \theta \) and change \( \tan \theta \) and \( \operatorname{cosec} \theta \) into \( \sin \theta \) and \( \cos \theta \). After combining and simplifying, we get \( \frac{1}{\cos^3 \theta} \), which is the same as \( (\sec \theta)^3 \). Then, we put in the value of \( \sec \theta \) we found at the start to get the answer. Second part: We need to find for which values of \( k \) this rule works. Since \( \tan^2 \theta \) cannot be a negative number, \( 1 - k^2 \) must be zero or positive. This means \( k^2 \) must be 1 or less. So, \( k \) can be any number from -1 to 1. This range makes sure all the math makes sense.

🎯 Exam Tip: When working with square roots or fractional exponents, always consider the domain restrictions. For expressions like \( \sqrt{A} \) or \( A^{3/2} \), the base \( A \) must be non-negative.

Question 9. If \( \sec \theta + \tan \theta = p \), obtain the values of \( \sec \theta \), \( \tan \theta \) and \( \sin \theta \) in terms of \( p \).
Answer:
We are given the equation \( \sec \theta + \tan \theta = p \) ...(1), and we need to find \( \sec \theta \), \( \tan \theta \), and \( \sin \theta \) in terms of \( p \).
We use the fundamental trigonometric identity: \( \sec^2 \theta - \tan^2 \theta = 1 \).
This identity can be factored using the difference of squares formula, \( A^2 - B^2 = (A+B)(A-B) \):
\( (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \)
Since we know \( \sec \theta + \tan \theta = p \), we can substitute this into the factored identity:
\( p(\sec \theta - \tan \theta) = 1 \)
From this, we find \( \sec \theta - \tan \theta \):
\( \sec \theta - \tan \theta = \frac{1}{p} \) ...(2)
Now we have a system of two simple linear equations with \( \sec \theta \) and \( \tan \theta \):
1. \( \sec \theta + \tan \theta = p \)
2. \( \sec \theta - \tan \theta = \frac{1}{p} \)
To find \( \sec \theta \), we add these two equations:
\( (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = p + \frac{1}{p} \)
\( \implies 2 \sec \theta = \frac{p^2 + 1}{p} \)
Dividing by 2, we get \( \sec \theta \):
\( \sec \theta = \frac{p^2 + 1}{2p} \)
To find \( \tan \theta \), we subtract Equation (2) from Equation (1):
\( (\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = p - \frac{1}{p} \)
\( \implies 2 \tan \theta = \frac{p^2 - 1}{p} \)
Dividing by 2, we get \( \tan \theta \):
\( \tan \theta = \frac{p^2 - 1}{2p} \)
Finally, to find \( \sin \theta \), we use the relationship \( \sin \theta = \frac{\tan \theta}{\sec \theta} \).
Substitute the expressions we found for \( \tan \theta \) and \( \sec \theta \):
\( \sin \theta = \frac{\frac{p^2 - 1}{2p}}{\frac{p^2 + 1}{2p}} \)
We can cancel out the common \( \frac{1}{2p} \) term from the numerator and denominator:
\( \sin \theta = \frac{p^2 - 1}{p^2 + 1} \)
This approach provides all required values in terms of \( p \). This method is a common way to solve for trigonometric functions when their sum or difference is given.
In simple words: We are given a rule with \( \sec \theta \), \( \tan \theta \), and \( p \). We also know a special math rule: \( \sec^2 \theta - \tan^2 \theta = 1 \). We can break this rule into two parts: \( (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \). Since we know \( \sec \theta + \tan \theta \) is \( p \), we can find that \( \sec \theta - \tan \theta \) is \( \frac{1}{p} \). Now we have two easy rules. We add them together to find \( \sec \theta \). Then we subtract them to find \( \tan \theta \). To find \( \sin \theta \), we just divide the value of \( \tan \theta \) by the value of \( \sec \theta \). This gives us all three answers using only \( p \).

🎯 Exam Tip: Always remember the fundamental identities like \( \sec^2 \theta - \tan^2 \theta = 1 \) and \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \), as they often simplify problems involving sums and differences.

Question 10. If \( \cot \theta(1 + \sin \theta) = 4m \) and \( \cot \theta (1 – \sin \theta) = 4n \) then prove that \( (m^2 – n^2)^2 = mn \).
Answer:
We are given two equations:
1. \( \cot \theta(1 + \sin \theta) = 4m \)
2. \( \cot \theta(1 - \sin \theta) = 4n \)
Our goal is to prove that \( (m^2 – n^2)^2 = mn \).
First, let's find expressions for \( m \) and \( n \) by dividing the given equations by 4:
\( m = \frac{\cot \theta(1 + \sin \theta)}{4} \)
\( n = \frac{\cot \theta(1 - \sin \theta)}{4} \)
Next, we calculate \( m^2 - n^2 \). We square the expressions for \( m \) and \( n \), then subtract:
\( m^2 = \frac{\cot^2 \theta (1 + \sin \theta)^2}{16} \)
\( n^2 = \frac{\cot^2 \theta (1 - \sin \theta)^2}{16} \)
\( m^2 - n^2 = \frac{\cot^2 \theta}{16} [(1 + \sin \theta)^2 - (1 - \sin \theta)^2] \)
We use the algebraic identity \( (a+b)^2 - (a-b)^2 = 4ab \). Applying this, \( (1 + \sin \theta)^2 - (1 - \sin \theta)^2 \) simplifies to \( 4(1)(\sin \theta) = 4 \sin \theta \).
So, \( m^2 - n^2 = \frac{\cot^2 \theta}{16} [4 \sin \theta] \)
\( \implies m^2 - n^2 = \frac{\cot^2 \theta \sin \theta}{4} \)
Now, we need to square this result to find the left-hand side (LHS) of the identity to be proven:
\( (m^2 - n^2)^2 = \left( \frac{\cot^2 \theta \sin \theta}{4} \right)^2 \)
\( \implies = \frac{\cot^4 \theta \sin^2 \theta}{16} \)
We can rewrite \( \cot^4 \theta \) as \( \left(\frac{\cos^2 \theta}{\sin^2 \theta}\right)^2 = \frac{\cos^4 \theta}{\sin^4 \theta} \):
\( \implies = \frac{\cos^4 \theta}{\sin^4 \theta} \times \frac{\sin^2 \theta}{16} \)
\( \implies (m^2 - n^2)^2 = \frac{\cos^4 \theta}{16 \sin^2 \theta} \) ...(A)
Next, let's calculate \( mn \) for the right-hand side (RHS) of the identity:
\( mn = \left( \frac{\cot \theta(1 + \sin \theta)}{4} \right) \left( \frac{\cot \theta(1 - \sin \theta)}{4} \right) \)
\( \implies = \frac{\cot^2 \theta (1 + \sin \theta)(1 - \sin \theta)}{16} \)
The term \( (1 + \sin \theta)(1 - \sin \theta) \) is a difference of squares, \( 1^2 - \sin^2 \theta = 1 - \sin^2 \theta \).
Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \):
\( mn = \frac{\cot^2 \theta \cos^2 \theta}{16} \)
Again, we replace \( \cot^2 \theta \) with \( \frac{\cos^2 \theta}{\sin^2 \theta} \):
\( \implies mn = \frac{\frac{\cos^2 \theta}{\sin^2 \theta} \times \cos^2 \theta}{16} \)
\( \implies mn = \frac{\cos^4 \theta}{16 \sin^2 \theta} \) ...(B)
By comparing the final expressions for \( (m^2 - n^2)^2 \) (A) and \( mn \) (B), we see they are identical.
Thus, \( (m^2 – n^2)^2 = mn \) is proven. This solution demonstrates using algebraic identities and trigonometric substitutions to simplify and prove complex equations.
In simple words: We are given two rules that give us values for \( m \) and \( n \). We need to show that a certain math rule for \( m \) and \( n \) is true. First, we find out what \( m^2 - n^2 \) is. We use a shortcut rule that \( (a+b)^2 - (a-b)^2 \) is \( 4ab \). This makes \( m^2 - n^2 \) much simpler. Then we square this result. Next, we multiply \( m \) and \( n \) together. We use another shortcut rule that \( (a+b)(a-b) \) is \( a^2 - b^2 \). We also use the basic rule that \( 1 - \sin^2 \theta \) is \( \cos^2 \theta \). After doing all the math, we find that both sides of the rule we need to prove are exactly the same. This shows the rule is true.

🎯 Exam Tip: Recognizing algebraic identities like \( (a+b)^2-(a-b)^2 = 4ab \) and \( (a+b)(a-b) = a^2-b^2 \) can significantly simplify the calculation steps in trigonometric proofs.

Question 11. If cosec \( \theta \) – sin \( \theta \) = \( a^3 \), sec \( \theta \) – cos \( \theta \) = \( b^3 \) then prove that \( a^2b^2(a^2 + b^2) = 1 \).
Answer:
Given:
\( a^3 = \text{cosec } \theta - \sin \theta \)
We know that \( \text{cosec } \theta = \frac{1}{\sin \theta} \). So, substitute this:
\( a^3 = \frac{1}{\sin \theta} - \sin \theta \)
\( a^3 = \frac{1 - \sin^2 \theta}{\sin \theta} \)
Since \( 1 - \sin^2 \theta = \cos^2 \theta \), we get:
\( a^3 = \frac{\cos^2 \theta}{\sin \theta} \)
This means \( a = \left( \frac{\cos^2 \theta}{\sin \theta} \right)^{1/3} \)
Then, \( a^2 = \left( \frac{\cos^2 \theta}{\sin \theta} \right)^{2/3} = \frac{\cos^{4/3} \theta}{\sin^{2/3} \theta} \)

Also given:
\( b^3 = \sec \theta - \cos \theta \)
We know that \( \sec \theta = \frac{1}{\cos \theta} \). So, substitute this:
\( b^3 = \frac{1}{\cos \theta} - \cos \theta \)
\( b^3 = \frac{1 - \cos^2 \theta}{\cos \theta} \)
Since \( 1 - \cos^2 \theta = \sin^2 \theta \), we get:
\( b^3 = \frac{\sin^2 \theta}{\cos \theta} \)
This means \( b = \left( \frac{\sin^2 \theta}{\cos \theta} \right)^{1/3} \)
Then, \( b^2 = \left( \frac{\sin^2 \theta}{\cos \theta} \right)^{2/3} = \frac{\sin^{4/3} \theta}{\cos^{2/3} \theta} \)

Now, let's find \( a^2 b^2 \):
\( a^2 b^2 = \left( \frac{\cos^{4/3} \theta}{\sin^{2/3} \theta} \right) \times \left( \frac{\sin^{4/3} \theta}{\cos^{2/3} \theta} \right) \)
\( a^2 b^2 = \cos^{(4/3 - 2/3)} \theta \times \sin^{(4/3 - 2/3)} \theta \)
\( a^2 b^2 = \cos^{2/3} \theta \times \sin^{2/3} \theta \)

Next, let's find \( a^2 + b^2 \):
\( a^2 + b^2 = \frac{\cos^{4/3} \theta}{\sin^{2/3} \theta} + \frac{\sin^{4/3} \theta}{\cos^{2/3} \theta} \)
To add these, we find a common denominator:
\( a^2 + b^2 = \frac{\cos^{4/3} \theta \cdot \cos^{2/3} \theta + \sin^{4/3} \theta \cdot \sin^{2/3} \theta}{\sin^{2/3} \theta \cdot \cos^{2/3} \theta} \)
\( a^2 + b^2 = \frac{\cos^{(4/3 + 2/3)} \theta + \sin^{(4/3 + 2/3)} \theta}{\sin^{2/3} \theta \cdot \cos^{2/3} \theta} \)
\( a^2 + b^2 = \frac{\cos^2 \theta + \sin^2 \theta}{\sin^{2/3} \theta \cdot \cos^{2/3} \theta} \)
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we get:
\( a^2 + b^2 = \frac{1}{\sin^{2/3} \theta \cdot \cos^{2/3} \theta} \)

Finally, multiply \( a^2 b^2 \) by \( (a^2 + b^2) \):
\( a^2 b^2 (a^2 + b^2) = (\cos^{2/3} \theta \cdot \sin^{2/3} \theta) \times \left( \frac{1}{\sin^{2/3} \theta \cdot \cos^{2/3} \theta} \right) \)
\( a^2 b^2 (a^2 + b^2) = 1 \)
This proves the identity. We combine the simplified forms of \( a^2 \) and \( b^2 \) to show that the final product equals one, using fundamental trigonometric identities.
In simple words: First, we change \( a^3 \) and \( b^3 \) into expressions using only sine and cosine. Then, we find \( a^2 \) and \( b^2 \). We multiply \( a^2 \) and \( b^2 \) together, and then add them. Finally, we multiply these two results to show that the answer is 1.

🎯 Exam Tip: When proving identities, always simplify each side separately or one part at a time using fundamental identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) until both sides match or the expression simplifies to the desired result.

Question 12. Eliminate \( \theta \) from the equations \( a \sec \theta - c \tan \theta = b \), \( b \sec \theta + d \tan \theta = c \).
Answer:
We are given two equations:
(1) \( a \sec \theta - c \tan \theta = b \)
(2) \( b \sec \theta + d \tan \theta = c \)

To eliminate \( \theta \), we first solve these equations for \( \sec \theta \) and \( \tan \theta \), then use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \).

Step 1: Eliminate \( \sec \theta \) to find \( \tan \theta \).
Multiply equation (1) by \( b \):
\( ab \sec \theta - bc \tan \theta = b^2 \) (Equation 3)
Multiply equation (2) by \( a \):
\( ab \sec \theta + ad \tan \theta = ac \) (Equation 4)
Subtract Equation 3 from Equation 4:
\( (ab \sec \theta + ad \tan \theta) - (ab \sec \theta - bc \tan \theta) = ac - b^2 \)
\( ab \sec \theta + ad \tan \theta - ab \sec \theta + bc \tan \theta = ac - b^2 \)
\( (ad + bc) \tan \theta = ac - b^2 \)
Solving for \( \tan \theta \):
\( \tan \theta = \frac{ac - b^2}{ad + bc} \) (Equation 5)

Step 2: Eliminate \( \tan \theta \) to find \( \sec \theta \).
Multiply equation (1) by \( d \):
\( ad \sec \theta - cd \tan \theta = bd \) (Equation 6)
Multiply equation (2) by \( c \):
\( bc \sec \theta + cd \tan \theta = c^2 \) (Equation 7)
Add Equation 6 and Equation 7:
\( (ad \sec \theta - cd \tan \theta) + (bc \sec \theta + cd \tan \theta) = bd + c^2 \)
\( ad \sec \theta + bc \sec \theta - cd \tan \theta + cd \tan \theta = bd + c^2 \)
\( (ad + bc) \sec \theta = c^2 + bd \)
Solving for \( \sec \theta \):
\( \sec \theta = \frac{c^2 + bd}{ad + bc} \) (Equation 8)

Step 3: Use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \).
Substitute the expressions for \( \sec \theta \) from Equation 8 and \( \tan \theta \) from Equation 5 into the identity:
\( \left( \frac{c^2 + bd}{ad + bc} \right)^2 - \left( \frac{ac - b^2}{ad + bc} \right)^2 = 1 \)
Since the denominators are the same, we can combine the fractions:
\( \frac{(c^2 + bd)^2 - (ac - b^2)^2}{(ad + bc)^2} = 1 \)
Multiply both sides by \( (ad + bc)^2 \):
\( (c^2 + bd)^2 - (ac - b^2)^2 = (ad + bc)^2 \)
This equation has no \( \theta \) and is the result after elimination. This problem shows how algebraic manipulation can isolate variables and reveal underlying relationships.
In simple words: We have two equations with \( \sec \theta \) and \( \tan \theta \). We solve these equations to find what \( \sec \theta \) equals and what \( \tan \theta \) equals, without \( \theta \) itself. Then, we use the rule that \( \sec^2 \theta - \tan^2 \theta \) is always 1. We put our solved values into this rule, and this removes \( \theta \) completely, giving us a final equation with only \( a, b, c, d \).

🎯 Exam Tip: When eliminating trigonometric functions, always aim to express them in terms of other variables (like \( \sec \theta \) and \( \tan \theta \) here) and then use a fundamental identity that connects those functions (like \( \sec^2 \theta - \tan^2 \theta = 1 \)).