Samacheer Kalvi Class 11 Business Maths Solutions Chapter 4 Trigonometry Exercise 4.1

Question 1. Convert the following degree measure into radian measure
(i) 60°
(ii) 150°
(iii) 240°
(iv) -320°
Answer: To convert degrees to radians, we multiply the degree measure by \( \frac{\pi}{180} \). This is because \( 180^\circ \) is equal to \( \pi \) radians.
(i) \( 60^\circ = 60 \times \frac{\pi}{180} \) radians \( = \frac{\pi}{3} \) radians.
(ii) \( 150^\circ = 150 \times \frac{\pi}{180} \) radians \( = \frac{5\pi}{6} \) radians.
(iii) \( 240^\circ = 240 \times \frac{\pi}{180} \) radians \( = \frac{4\pi}{3} \) radians.
(iv) \( -320^\circ = -320 \times \frac{\pi}{180} \) radians \( = \frac{-16\pi}{9} \) radians.
In simple words: To change a degree number into radians, just multiply it by \( \frac{\pi}{180} \). This calculation gives you the same angle in a different unit, using pi instead of degrees.

🎯 Exam Tip: Remember the conversion factor: \( 1^\circ = \frac{\pi}{180} \) radians. Always simplify the fraction to its lowest terms.

Question 2. Find the degree measure corresponding to the following radian measure.
(i) \( \frac{\pi}{8} \)
(ii) \( \frac{9 \pi}{5} \)
(iii) -3
(iv) \( \frac{11 \pi}{18} \)
Answer: To convert radians to degrees, we multiply the radian measure by \( \frac{180^\circ}{\pi} \). This is the inverse of converting degrees to radians.
(i) \( \frac{\pi}{8} \) radians \( = \frac{\pi}{8} \times \frac{180^\circ}{\pi} = \frac{180^\circ}{8} = 22.5^\circ \)
Since \( 0.5^\circ = (0.5 \times 60)' = 30' \), we can write \( 22.5^\circ \) as \( 22^\circ 30' \).
(ii) \( \frac{9\pi}{5} \) radians \( = \frac{9\pi}{5} \times \frac{180^\circ}{\pi} = 9 \times 36^\circ = 324^\circ \).
(iii) -3 radians \( = -3 \times \frac{180^\circ}{\pi} = -3 \times \frac{180^\circ}{(22/7)} = -3 \times \frac{180^\circ \times 7}{22} = - \frac{180 \times 21}{22} = - \frac{90 \times 21}{11} = - \frac{1890}{11} \approx -171.81^\circ \)
To convert \( 0.81^\circ \) to minutes: \( 0.81 \times 60' \approx 48.6' \). So, \( -171.81^\circ \) is approximately \( -171^\circ 48' \).
(iv) \( \frac{11\pi}{18} \) radians \( = \frac{11\pi}{18} \times \frac{180^\circ}{\pi} = 11 \times 10^\circ = 110^\circ \).
In simple words: To change a radian number back into degrees, you multiply it by \( \frac{180}{\pi} \). This reverses the first conversion, giving you the angle in degrees.

🎯 Exam Tip: For accuracy, especially with decimals in degrees, remember to convert the decimal part of the degree into minutes by multiplying by 60.

Question 3. Determine the quadrants in which the following degree lie.
(i) 380°
(ii) -140°
(iii) 1195°
Answer: Angles are measured counter-clockwise from the positive x-axis for positive values, and clockwise for negative values. Each full circle is \( 360^\circ \).
(i) For \( 380^\circ \): We can write \( 380^\circ = 360^\circ + 20^\circ \). After one full rotation, the angle is \( 20^\circ \), which lies in the first quadrant.
(ii) For \( -140^\circ \): A negative angle means moving clockwise. From \( 0^\circ \), moving clockwise \( 90^\circ \) reaches the negative y-axis. Moving another \( 50^\circ \) (total \( 90^\circ + 50^\circ = 140^\circ \)) places the angle in the third quadrant.
(iii) For \( 1195^\circ \): We can divide \( 1195 \) by \( 360 \) to find the number of full rotations: \( 1195 = 3 \times 360 + 115 \). So, after three full rotations, the angle is \( 115^\circ \). An angle of \( 115^\circ \) lies between \( 90^\circ \) and \( 180^\circ \), which is the second quadrant.
In simple words: To find the quadrant, take away full circles (360 degrees) until you have an angle between 0 and 360 degrees. Then see which quarter-section (quadrant) that angle falls into.

🎯 Exam Tip: Always subtract multiples of \( 360^\circ \) for positive angles or add multiples of \( 360^\circ \) for negative angles to get an equivalent angle between \( 0^\circ \) and \( 360^\circ \) before determining the quadrant.

Question 4. Find the values of each of the following trigonometric ratios.
(i) sin 300°
(ii) cos(-210°)
(iii) sec 390°
(iv) tan(-855°)
(v) cosec 1125°
Answer: We use reduction formulas and periodicity of trigonometric functions to find these values. The signs depend on the quadrant the angle lies in.
(i) \( \sin 300^\circ = \sin(360^\circ - 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2} \). ( \( 300^\circ \) is in the 4th quadrant, where sine is negative).
(ii) \( \cos(-210^\circ) = \cos(210^\circ) \) (since \( \cos(-\theta) = \cos \theta \)).
\( \cos(210^\circ) = \cos(180^\circ + 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \). ( \( 210^\circ \) is in the 3rd quadrant, where cosine is negative).
(iii) \( \sec 390^\circ = \sec(360^\circ + 30^\circ) = \sec 30^\circ \).
\( \sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{(\sqrt{3}/2)} = \frac{2}{\sqrt{3}} \).
(iv) \( \tan(-855^\circ) = -\tan(855^\circ) \) (since \( \tan(-\theta) = -\tan \theta \)).
\( 855^\circ = 2 \times 360^\circ + 135^\circ \). So \( -\tan(855^\circ) = -\tan(135^\circ) \).
\( -\tan(135^\circ) = -\tan(180^\circ - 45^\circ) = -(-\tan 45^\circ) = \tan 45^\circ = 1 \).
(v) \( \operatorname{cosec} 1125^\circ = \operatorname{cosec}(3 \times 360^\circ + 45^\circ) = \operatorname{cosec} 45^\circ \).
\( \operatorname{cosec} 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{(1/\sqrt{2})} = \sqrt{2} \).
In simple words: To find these values, first adjust the angle by adding or subtracting multiples of 360 degrees until it's between 0 and 360 degrees. Then, use special angle rules and quadrant signs to find the exact value.

🎯 Exam Tip: Memorize the basic trigonometric values for \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \) and understand the sign changes in each quadrant. This helps reduce any angle to a principal value.

Question 5. Prove that:
(i) \( \tan(-225^\circ) \cot(-405^\circ) - \tan(-765^\circ) \cot(675^\circ) = 0 \).
(ii) \( 2 \sin^2 \frac{\pi}{6} + \operatorname{cosec}^2 \frac{7 \pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2} \).
(iii) \( \sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)=-1 \).
Answer: We will evaluate each term separately using trigonometric identities and then substitute them into the given expressions to prove the identities.
(i) Let's evaluate each term:
\( \tan(-225^\circ) = -\tan(225^\circ) = -\tan(180^\circ + 45^\circ) = -(\tan 45^\circ) = -1 \).
\( \cot(-405^\circ) = -\cot(405^\circ) = -\cot(360^\circ + 45^\circ) = -\cot 45^\circ = -1 \).
\( \tan(-765^\circ) = -\tan(765^\circ) = -\tan(2 \times 360^\circ + 45^\circ) = -\tan 45^\circ = -1 \).
\( \cot(675^\circ) = \cot(360^\circ + 315^\circ) = \cot(315^\circ) = \cot(360^\circ - 45^\circ) = -\cot 45^\circ = -1 \).
Now substitute these values into the LHS:
LHS \( = (-1)(-1) - (-1)(-1) = 1 - 1 = 0 \). This equals the RHS. Hence proved.

(ii) Let's evaluate the LHS:
LHS \( = 2 \sin^2 \frac{\pi}{6} + \operatorname{cosec}^2 \frac{7 \pi}{6} \cos^2 \frac{\pi}{3} \).
We know \( \sin \frac{\pi}{6} = \sin 30^\circ = \frac{1}{2} \).
And \( \cos \frac{\pi}{3} = \cos 60^\circ = \frac{1}{2} \).
For \( \operatorname{cosec} \frac{7 \pi}{6} \): \( \frac{7 \pi}{6} = \pi + \frac{\pi}{6} \), which is \( 180^\circ + 30^\circ \). This angle is in the 3rd quadrant where cosec is negative.
So, \( \operatorname{cosec} \frac{7 \pi}{6} = \operatorname{cosec}(180^\circ + 30^\circ) = -\operatorname{cosec} 30^\circ = -2 \).
Now substitute these values into the LHS:
LHS \( = 2 \left(\frac{1}{2}\right)^2 + (-2)^2 \left(\frac{1}{2}\right)^2 \)
\( = 2 \left(\frac{1}{4}\right) + 4 \left(\frac{1}{4}\right) \)
\( = \frac{2}{4} + \frac{4}{4} = \frac{6}{4} = \frac{3}{2} \). This equals the RHS. Hence proved.

(iii) Let's evaluate each term for LHS:
\( \sec \left(\frac{3 \pi}{2}-\theta\right) = \sec(270^\circ - \theta) = -\operatorname{cosec} \theta \).
\( \sec \left(\theta-\frac{5 \pi}{2}\right) = \sec \left(-\left(\frac{5 \pi}{2}-\theta\right)\right) = \sec \left(\frac{5 \pi}{2}-\theta\right) \) (since \( \sec(-\phi) = \sec \phi \)).
\( \sec \left(\frac{5 \pi}{2}-\theta\right) = \sec(450^\circ - \theta) = \sec(360^\circ + 90^\circ - \theta) = \sec(90^\circ - \theta) = \operatorname{cosec} \theta \).
\( \tan \left(\frac{5 \pi}{2}+\theta\right) = \tan(450^\circ + \theta) = \tan(360^\circ + 90^\circ + \theta) = \tan(90^\circ + \theta) = -\cot \theta \).
\( \tan \left(\theta-\frac{5 \pi}{2}\right) = \tan \left(-\left(\frac{5 \pi}{2}-\theta\right)\right) = -\tan \left(\frac{5 \pi}{2}-\theta\right) \).
\( -\tan \left(\frac{5 \pi}{2}-\theta\right) = -\tan(450^\circ - \theta) = -\tan(360^\circ + 90^\circ - \theta) = -\tan(90^\circ - \theta) = -\cot \theta \).
Now substitute these into the LHS:
LHS \( = (-\operatorname{cosec} \theta)(\operatorname{cosec} \theta) + (-\cot \theta)(-\cot \theta) \)
\( = -\operatorname{cosec}^2 \theta + \cot^2 \theta \).
We know the identity \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \), so \( \cot^2 \theta = \operatorname{cosec}^2 \theta - 1 \).
LHS \( = -\operatorname{cosec}^2 \theta + (\operatorname{cosec}^2 \theta - 1) = -1 \). This equals the RHS. Hence proved.
In simple words: For proving these math statements, break them into smaller parts. Calculate each part using basic angle rules and quadrant signs. Then put them back together to show that one side equals the other side.

🎯 Exam Tip: When proving identities, always try to simplify one side (usually the LHS) until it matches the other side (RHS). Remember common identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \).

Question 6. If A, B, C, D are angles of a cyclic quadrilateral, prove that: cos A + cos B + cos C + cos D = 0.
Answer: A cyclic quadrilateral is a four-sided shape whose vertices all lie on a single circle. A key property of a cyclic quadrilateral is that the sum of its opposite angles is \( 180^\circ \).
Given A, B, C, D are angles of a cyclic quadrilateral, we have:
\( A + C = 180^\circ \implies C = 180^\circ - A \).
\( B + D = 180^\circ \implies D = 180^\circ - B \).
Now consider the Left Hand Side (LHS) of the equation to be proved:
LHS \( = \cos A + \cos B + \cos C + \cos D \).
Substitute the expressions for C and D:
LHS \( = \cos A + \cos B + \cos(180^\circ - A) + \cos(180^\circ - B) \).
Using the identity \( \cos(180^\circ - x) = -\cos x \):
LHS \( = \cos A + \cos B - \cos A - \cos B \).
LHS \( = 0 \).
Since LHS equals RHS, the statement is proved.
In simple words: For a cyclic quadrilateral, opposite angles add up to 180 degrees. If you use this fact and the cosine rule that \( \cos(180^\circ - A) = -\cos A \), then all the cosine terms will cancel each other out, making the total sum zero.

🎯 Exam Tip: The crucial step here is to remember the property of cyclic quadrilaterals (opposite angles sum to \( 180^\circ \)) and the trigonometric identity for \( \cos(180^\circ - x) \).

Question 7. Prove that
(i) \( \frac{\sin(180^\circ - \theta) \cos(90^\circ + \theta)\tan(270^\circ - \theta) \cot(360^\circ - \theta)}{\sin(360^\circ - \theta) \cos(360^\circ + \theta)\sin(270^\circ - \theta) \operatorname{cosec}(-\theta)} = -1 \).
(ii) \( \sin \theta \cdot \cos\left(\sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta\right) = 1 \).
Answer: We will simplify the LHS of each equation using trigonometric identities and quadrant rules to show it equals the RHS.
(i) Let's simplify each term in the numerator and denominator:
Numerator:
\( \sin(180^\circ - \theta) = \sin \theta \)
\( \cos(90^\circ + \theta) = -\sin \theta \)
\( \tan(270^\circ - \theta) = \cot \theta \)
\( \cot(360^\circ - \theta) = -\cot \theta \)
Product of numerator terms \( = (\sin \theta)(-\sin \theta)(\cot \theta)(-\cot \theta) = \sin^2 \theta \cot^2 \theta \).
Denominator:
\( \sin(360^\circ - \theta) = -\sin \theta \)
\( \cos(360^\circ + \theta) = \cos \theta \)
\( \sin(270^\circ - \theta) = -\cos \theta \)
\( \operatorname{cosec}(-\theta) = -\operatorname{cosec} \theta \)
Product of denominator terms \( = (-\sin \theta)(\cos \theta)(-\cos \theta)(-\operatorname{cosec} \theta) = -\sin \theta \cos^2 \theta \operatorname{cosec} \theta \).
Now substitute these back into the expression:
LHS \( = \frac{\sin^2 \theta \cot^2 \theta}{-\sin \theta \cos^2 \theta \operatorname{cosec} \theta} \)
Replace \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \):
LHS \( = \frac{\sin^2 \theta \left(\frac{\cos^2 \theta}{\sin^2 \theta}\right)}{-\sin \theta \cos^2 \theta \left(\frac{1}{\sin \theta}\right)} \)
LHS \( = \frac{\cos^2 \theta}{-\cos^2 \theta} = -1 \). This equals the RHS. Hence proved.

(ii) Let's simplify the LHS:
LHS \( = \sin \theta \cdot \cos\left(\sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta\right) \).
First, simplify the terms inside the parenthesis:
\( \sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta \).
\( \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta \).
So the expression inside the parenthesis becomes:
\( \cos \theta \cdot \operatorname{cosec} \theta + \sin \theta \cdot \sec \theta \)
\( = \cos \theta \cdot \frac{1}{\sin \theta} + \sin \theta \cdot \frac{1}{\cos \theta} \)
\( = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \)
\( = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \).
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), this becomes \( \frac{1}{\sin \theta \cos \theta} \).
Now substitute this back into the LHS:
LHS \( = \sin \theta \cdot \left( \frac{1}{\sin \theta \cos \theta} \right) = \frac{1}{\cos \theta} = \sec \theta \).
Wait, there is a typo in the question, the LHS cannot become \( \cos(X) \), it is \( \sin \theta \cdot (\ldots) \) in the original question.
The problem in the OCR text appears to have a structure issue for Q7(ii). It reads \( \sin \theta \cdot \cos\{\ldots\} \), but the solution provided for this part of the question actually simplifies \( \sin \theta \cdot \operatorname{cosec} \theta + \cos \theta \cdot \sec \theta \) from within the curly braces and shows the LHS as \( \cos^2 \theta + \sin^2 \theta = 1 \). This implies the question might have intended something like \( \sin \theta \cdot (\sin(\frac{\pi}{2} - \theta) \cdot \operatorname{cosec} \theta + \cos(\frac{\pi}{2} - \theta) \cdot \sec \theta) \) to be equal to 1. Given the OCR, I will assume the provided solution's intention was to prove the expression inside the curly braces to be 1, multiplied by sin theta, or that it was missing a term. However, following the given solution which ends with \( \cos^2 \theta + \sin^2 \theta = 1 \), I will reproduce the steps as shown. Let's re-examine the OCR for (ii): \( \sin \theta \cdot \cos\{\sin(\frac{\pi}{2} – Ө) \cdot \operatorname{cosec} \theta + \cos(\frac{\pi}{2} – Ө) \cdot \sec \theta\} = 1 \). The solution starts with LHS \( = \sin \theta \cos\{\sin(\frac{\pi}{2} - \theta) \cdot \operatorname{cosec} \theta + \cos(\frac{\pi}{2} - \theta) \cdot \sec \theta\} \). Then it says: LHS \( = \sin \theta \cos\{\cos \theta \cdot \frac{1}{\sin \theta} + \sin \theta \cdot \frac{1}{\cos \theta}\} \) which is \( \sin \theta \cos\{\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}\} \). This simplifies to \( \sin \theta \cos\{\frac{1}{\sin \theta \cos \theta}\} = \sin \theta \cdot \cos(\operatorname{cosec} \theta \sec \theta) \). This cannot simplify to 1 as written. However, the provided solution for 7(ii) then calculates: \( = \sin \theta \cos \theta \{\frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta}\} \). This seems to be a different expression from the original question text. Let's follow the OCR's provided *solution* steps exactly: LHS \( = \sin \theta \cos \theta \left\{ \sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta \right\} \) This is actually what the solution text for 7(ii) implies, not what is explicitly written in the question line. I must stick to the question as written. Let's assume the question meant: \( \sin \theta \left( \sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta \right) = 1 \). (If the 'cos' was a typo and should be part of the expression inside the parenthesis). If it is: \( \sin \theta \cdot \left[ \left(\cos \theta \right) \cdot \left( \frac{1}{\sin \theta} \right) + \left(\sin \theta \right) \cdot \left( \frac{1}{\cos \theta} \right) \right] \) \( = \sin \theta \left[ \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right] \) \( = \sin \theta \left[ \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right] \) \( = \sin \theta \left[ \frac{1}{\sin \theta \cos \theta} \right] \) \( = \frac{1}{\cos \theta} = \sec \theta \). This doesn't equal 1. Given the actual OCR output for the solution for 7(ii) on page 10: LHS \( = \sin \theta \cos \theta \left\{ \sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta \right\} \) This is the expression the solution is actually evaluating, not what is in the question. If we use this expression: LHS \( = \sin \theta \cos \theta \left\{ (\cos \theta) \cdot (\frac{1}{\sin \theta}) + (\sin \theta) \cdot (\frac{1}{\cos \theta}) \right\} \) \( = \sin \theta \cos \theta \left\{ \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right\} \) \( = \sin \theta \cos \theta \left\{ \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right\} \) \( = \sin \theta \cos \theta \left\{ \frac{1}{\sin \theta \cos \theta} \right\} \) \( = 1 \). This matches the RHS. So, the question text in the PDF seems to be an error, and the solution provides the steps for proving: \( \sin \theta \cos \theta \left\{ \sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta \right\} = 1 \). I will present the solution for this expression, as it's what the source's "solution" section attempts to prove to be 1. (ii) We need to prove \( \sin \theta \cos \theta \left\{ \sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta \right\} = 1 \). Let's simplify the LHS:
LHS \( = \sin \theta \cos \theta \left\{ \sin\left(\frac{\pi}{2} - \theta\right) \cdot \operatorname{cosec} \theta + \cos\left(\frac{\pi}{2} - \theta\right) \cdot \sec \theta \right\} \).
Using the identities: \( \sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta \) and \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta \).
Also, \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \).
Substitute these into the expression:
LHS \( = \sin \theta \cos \theta \left\{ (\cos \theta) \cdot \left(\frac{1}{\sin \theta}\right) + (\sin \theta) \cdot \left(\frac{1}{\cos \theta}\right) \right\} \)
LHS \( = \sin \theta \cos \theta \left\{ \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right\} \)
Combine the terms inside the curly braces with a common denominator:
LHS \( = \sin \theta \cos \theta \left\{ \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right\} \)
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), the expression becomes:
LHS \( = \sin \theta \cos \theta \left\{ \frac{1}{\sin \theta \cos \theta} \right\} \)
LHS \( = 1 \). This equals the RHS. Hence proved.
In simple words: This proof uses basic trig rules like changing angles (e.g., sin(90-theta) becomes cos theta) and inverse definitions (e.g., cosec theta is 1/sin theta). After changing all parts, you find a common factor that cancels out, leaving just 1.

🎯 Exam Tip: When faced with complex expressions, simplify each trigonometric term using quadrant rules and identities first. Look for opportunities to use fundamental identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 8. Prove that: \( \cos 510^\circ \cos 330^\circ + \sin 390^\circ \cos 120^\circ = -1 \).
Answer: We need to evaluate the trigonometric ratios for each angle and then substitute them into the expression.
Let's simplify each term:
\( \cos 510^\circ = \cos(360^\circ + 150^\circ) = \cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \).
\( \cos 330^\circ = \cos(360^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2} \).
\( \sin 390^\circ = \sin(360^\circ + 30^\circ) = \sin 30^\circ = \frac{1}{2} \).
\( \cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \).
Now substitute these values into the LHS:
LHS \( = \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(-\frac{1}{2}\right) \)
LHS \( = -\frac{3}{4} - \frac{1}{4} \)
LHS \( = \frac{-3-1}{4} = \frac{-4}{4} = -1 \).
This equals the RHS. Hence proved.
In simple words: First, find the values of each sine and cosine part by simplifying the angles (like 510 degrees is same as 150 degrees). Then put these values into the equation and do the math to show that the left side equals -1.

🎯 Exam Tip: When dealing with angles greater than \( 360^\circ \) or negative angles, always reduce them to their equivalent angles between \( 0^\circ \) and \( 360^\circ \) using the periodicity of trigonometric functions.

Question 9. Prove that:
(i) \( \tan(\pi + x) \cot(x - \pi) - \cos(2\pi - x) \cos(2\pi + x) = \sin^2 x \).
(ii) \( \frac{\sin(180^\circ+A)\cos(90^\circ - A)\tan(270^\circ - A)}{\sec(540^\circ-A)\cos(360^\circ + A) \operatorname{cosec}(270^\circ + A)} = -\sin A \cos^2 A \).
Answer: We will simplify the LHS of each identity using trigonometric identities and reduction formulas to match the RHS.
(i) Let's simplify the LHS:
LHS \( = \tan(\pi + x) \cot(x - \pi) - \cos(2\pi - x) \cos(2\pi + x) \).
Using identities:
\( \tan(\pi + x) = \tan x \).
\( \cot(x - \pi) = \cot(-( \pi - x)) = -\cot(\pi - x) = -(-\cot x) = \cot x \).
\( \cos(2\pi - x) = \cos x \).
\( \cos(2\pi + x) = \cos x \).
Substitute these into the LHS:
LHS \( = (\tan x)(\cot x) - (\cos x)(\cos x) \)
LHS \( = 1 - \cos^2 x \).
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we have \( \sin^2 x = 1 - \cos^2 x \).
LHS \( = \sin^2 x \). This equals the RHS. Hence proved.

(ii) Let's simplify each term in the numerator and denominator:
Numerator:
\( \sin(180^\circ + A) = -\sin A \).
\( \cos(90^\circ - A) = \sin A \).
\( \tan(270^\circ - A) = \cot A \).
Product of numerator terms \( = (-\sin A)(\sin A)(\cot A) = -\sin^2 A \cot A \).
Denominator:
\( \sec(540^\circ - A) = \sec(360^\circ + 180^\circ - A) = \sec(180^\circ - A) = -\sec A \).
\( \cos(360^\circ + A) = \cos A \).
\( \operatorname{cosec}(270^\circ + A) = -\sec A \).
Product of denominator terms \( = (-\sec A)(\cos A)(-\sec A) \).
Since \( \sec A = \frac{1}{\cos A} \), this becomes \( \left(-\frac{1}{\cos A}\right)(\cos A)\left(-\frac{1}{\cos A}\right) = (-1)\left(-\frac{1}{\cos A}\right) = \frac{1}{\cos A} \).
Now substitute the simplified numerator and denominator back into the expression:
LHS \( = \frac{-\sin^2 A \cot A}{1/\cos A} \).
Replace \( \cot A = \frac{\cos A}{\sin A} \):
LHS \( = \frac{-\sin^2 A \left(\frac{\cos A}{\sin A}\right)}{1/\cos A} \)
LHS \( = \frac{-\sin A \cos A}{1/\cos A} \)
LHS \( = -\sin A \cos A \times \cos A = -\sin A \cos^2 A \).
This equals the RHS. Hence proved.
In simple words: For both parts, simplify each trigonometric function by adjusting the angles to common forms like \( (180+x) \) or \( (360-x) \). Then, use the simple definitions (like tan is 1/cot) and basic identities (like \( \sin^2 x + \cos^2 x = 1 \)) to make the left side equal the right side.

🎯 Exam Tip: Pay close attention to the signs of trigonometric functions in different quadrants. Angles like \( (\pi+x) \) or \( (270^\circ-A) \) affect both the function (e.g., tan to cot) and its sign.

Question 10. If \( \sin \theta = \frac{3}{5} \), \( \tan \phi = \frac{1}{2} \) and \( \frac{\pi}{2} < \theta < \pi < \phi < \frac{3 \pi}{2} \), then find the value of \( 8 \tan \theta - \sqrt{5} \sec \phi \).
Answer: We are given \( \sin \theta = \frac{3}{5} \) and that \( \theta \) is in the second quadrant (\( \frac{\pi}{2} < \theta < \pi \)).
In the second quadrant, sine is positive, cosine is negative, and tangent is negative.
We can find \( \cos \theta \) using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \)
\( \frac{9}{25} + \cos^2 \theta = 1 \)
\( \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \)
Since \( \theta \) is in the second quadrant, \( \cos \theta = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \).
Now, \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{-4/5} = -\frac{3}{4} \).

Next, we are given \( \tan \phi = \frac{1}{2} \) and that \( \phi \) is in the third quadrant (\( \pi < \phi < \frac{3 \pi}{2} \)).
In the third quadrant, tangent is positive, but cosine and secant are negative.
We can find \( \sec \phi \) using the identity \( 1 + \tan^2 \phi = \sec^2 \phi \).
\( 1 + \left(\frac{1}{2}\right)^2 = \sec^2 \phi \)
\( 1 + \frac{1}{4} = \sec^2 \phi \)
\( \sec^2 \phi = \frac{5}{4} \)
Since \( \phi \) is in the third quadrant, \( \sec \phi = -\sqrt{\frac{5}{4}} = -\frac{\sqrt{5}}{2} \).

Now we need to find the value of \( 8 \tan \theta - \sqrt{5} \sec \phi \).
Substitute the values we found:
\( 8 \left(-\frac{3}{4}\right) - \sqrt{5} \left(-\frac{\sqrt{5}}{2}\right) \)
\( = 2(-3) + \frac{5}{2} \)
\( = -6 + \frac{5}{2} \)
To add these, find a common denominator:
\( = \frac{-12}{2} + \frac{5}{2} = \frac{-12 + 5}{2} = -\frac{7}{2} \).
In simple words: First, find the missing trigonometric values (like cos and sec) for angles theta and phi, making sure to use the correct positive or negative sign for their given quadrants. Then, put these numbers into the expression and calculate the final answer.

🎯 Exam Tip: Always pay attention to the given quadrant for each angle. This determines the sign of the trigonometric ratios, which is crucial for the correct answer.