CBSE Class 11 Mathematics Trigonometric Functions Worksheet Set 09

Question 1. The difference between two acute angles of a right triangle is \(\frac{\pi}{9}\). Find the angles in degree.
Answer: Let the two acute angles of the right-angled triangle be \(x\) and \(y\).
In a right-angled triangle, the sum of the two acute angles is \(90^\circ\) (or \(\frac{\pi}{2}\) radians).
So, we have the system of equations:
\(x + y = 90^\circ\)
\(x - y = \frac{\pi}{9} \text{ radians}\)
Converting \(\frac{\pi}{9}\) radians to degrees:
\(\frac{\pi}{9} \times \frac{180^\circ}{\pi} = 20^\circ\)
Thus, \(x - y = 20^\circ\)
Adding both equations:
\((x + y) + (x - y) = 90^\circ + 20^\circ\)
\(2x = 110^\circ\)
\(x = 55^\circ\)
Substituting \(x = 55^\circ\) into \(x + y = 90^\circ\):
\(55^\circ + y = 90^\circ \implies y = 35^\circ\)
Therefore, the angles are \(55^\circ\) and \(35^\circ\).
In simple words: The two acute angles add up to 90 degrees, and their difference is given as 20 degrees. Solving these two simple equations gives the angles.

Exam Tip: Always convert radian measures to degrees first if the final answer is required in degrees.

Question 2. A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describe 88 metres when it was traced out \(72^\circ\) at the centre, find the length of the rope.
Answer: Let the length of the rope be \(r\), which acts as the radius of the circular path.
The arc length \(s\) is given as \(88\) metres.
The central angle \(\theta = 72^\circ\).
Converting \(\theta\) to radians:
\(\theta = 72^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{5} \text{ radians}\)
The formula relating arc length, radius, and angle in radians is:
\(s = r\theta\)
Substitute the known values into the formula:
\(88 = r \times \frac{2\pi}{5}\)
Using \(\pi = \frac{22}{7}\):
\(88 = r \times \frac{2 \times 22}{7 \times 5}\)
\(88 = r \times \frac{44}{35}\)
\(r = \frac{88 \times 35}{44}\)
\(r = 2 \times 35 = 70 \text{ metres}\)
Therefore, the length of the rope is \(70\) metres.
In simple words: Use the arc length formula \(s = r\theta\) after converting the angle from degrees to radians to find the radius (length of the rope).

Exam Tip: The formula \(s = r\theta\) requires \(\theta\) to be strictly in radians, not degrees.

Question 3. The angles of a triangle are A.P. such that the greatest is 5 times the least. Find the angles in radians.
Answer: Let the three angles of the triangle in Arithmetic Progression (A.P.) be \(a - d\), \(a\), and \(a + d\).
The sum of angles in a triangle is \(180^\circ\) (or \(\pi\) radians):
\((a - d) + a + (a + d) = \pi\)
\(3a = \pi \implies a = \frac{\pi}{3}\)
According to the question, the greatest angle is 5 times the least angle:
\(a + d = 5(a - d)\)
\(a + d = 5a - 5d\)
\(6d = 4a \implies d = \frac{2}{3}a\)
Substitute \(a = \frac{\pi}{3}\):
\(d = \frac{2}{3} \left(\frac{\pi}{3}\right) = \frac{2\pi}{9}\)
Now, find the three angles:
Least angle: \(a - d = \frac{\pi}{3} - \frac{2\pi}{9} = \frac{3\pi - 2\pi}{9} = \frac{\pi}{9}\)
Middle angle: \(a = \frac{\pi}{3} = \frac{3\pi}{9}\)
Greatest angle: \(a + d = \frac{\pi}{3} + \frac{2\pi}{9} = \frac{3\pi + 2\pi}{9} = \frac{5\pi}{9}\)
Therefore, the angles in radians are \(\frac{\pi}{9}\), \(\frac{3\pi}{9}\), and \(\frac{5\pi}{9}\).
In simple words: Assume the angles as \(a-d\), \(a\), and \(a+d\). Use the triangle angle sum property and the given condition to find \(a\) and \(d\).

Exam Tip: Representing A.P. terms as \(a-d\), \(a\), and \(a+d\) simplifies the sum instantly because the common difference \(d\) cancels out.

Question 4. Prove that (i) \((1 + \tan\alpha \tan\beta)^2 + (\tan\alpha - \tan\beta)^2 = \sec^2\alpha \sec^2\beta\)
Answer: Consider the Left Hand Side (LHS):
\(\text{LHS} = (1 + \tan\alpha \tan\beta)^2 + (\tan\alpha - \tan\beta)^2\)
Expand both terms using algebraic identities \((a+b)^2 = a^2 + b^2 + 2ab\) and \((a-b)^2 = a^2 + b^2 - 2ab\):
\(= 1 + \tan^2\alpha \tan^2\beta + 2\tan\alpha \tan\beta + \tan^2\alpha + \tan^2\beta - 2\tan\alpha \tan\beta\)
Cancel the term \(+2\tan\alpha \tan\beta\) and \(-2\tan\alpha \tan\beta\):
\(= 1 + \tan^2\alpha \tan^2\beta + \tan^2\alpha + \tan^2\beta\)
Group the terms by factoring:
\(= 1 + \tan^2\alpha + \tan^2\beta(1 + \tan^2\alpha)\)
\(= (1 + \tan^2\alpha)(1 + \tan^2\beta)\)
Using the trigonometric identity \(1 + \tan^2\theta = \sec^2\theta\):
\(= \sec^2\alpha \sec^2\beta = \text{RHS}\)
Hence proved.
In simple words: Expand the squares, cancel out the middle terms, and factor by grouping to convert tans into secants.

Exam Tip: Recognizing standard algebraic expansion followed by grouping is crucial for trigonometric proofs.

Question 4. (ii) \(\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \sec A \csc A + 1\)
Answer: Consider the Left Hand Side (LHS):
\(\text{LHS} = \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}\)
Convert all trigonometric ratios into sine and cosine: \(\tan A = \frac{\sin A}{\cos A}\) and \(\cot A = \frac{\cos A}{\sin A}\):
\(= \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}\)
Simplify the denominators:
\(= \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}}\)
\(= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} + \frac{\cos^2 A}{\sin A(\cos A - \sin A)}\)
Change the second denominator to match \((\sin A - \cos A)\) by factoring out a negative sign:
\(= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A)}\)
Take the common denominator \(\sin A \cos A (\sin A - \cos A)\):
\(= \frac{\sin^3 A - \cos^3 A}{\sin A \cos A (\sin A - \cos A)}\)
Apply the algebraic formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\):
\(= \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A \cos A (\sin A - \cos A)}\)
Cancel the common term \((\sin A - \cos A)\):
\(= \frac{\sin^2 A + \cos^2 A + \sin A \cos A}{\sin A \cos A}\)
Since \(\sin^2 A + \cos^2 A = 1\):
\(= \frac{1 + \sin A \cos A}{\sin A \cos A}\)
\(= \frac{1}{\sin A \cos A} + \frac{\sin A \cos A}{\sin A \cos A}\)
\(= \sec A \csc A + 1 = \text{RHS}\)
Hence proved.
In simple words: Convert everything to sine and cosine, find a common denominator using the difference of cubes formula, and simplify.

Exam Tip: When stuck in trigonometric identities involving tan and cot, converting them to sine and cosine is a reliable fallback strategy.

Question 5. If \(\sin\theta = \frac{12}{13}\) and \(\theta\) lies in 2nd quadrant, then find the value of \(8\tan\theta - 5\sec\theta\)
Answer: Given \(\sin\theta = \frac{12}{13}\) and \(\theta\) lies in the second quadrant.
In the second quadrant, sine is positive, but cosine, secant, tangent, and cotangent are negative.
Find \(\cos\theta\) using the identity \(\cos^2\theta = 1 - \sin^2\theta\):
\(\cos\theta = -\sqrt{1 - \sin^2\theta} = -\sqrt{1 - \left(\frac{12}{13}\right)^2}\)
\(\cos\theta = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}\)
Now, find \(\tan\theta\) and \(\sec\theta\):
\(\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5}\)
\(\sec\theta = \frac{1}{\cos\theta} = \frac{1}{-\frac{5}{13}} = -\frac{13}{5}\)
Substitute these values into the expression \(8\tan\theta - 5\sec\theta\):
\(= 8\left(-\frac{12}{5}\right) - 5\left(-\frac{13}{5}\right)\)
\(= -\frac{96}{5} + \frac{65}{5}\)
\(= \frac{-96 + 65}{5} = -\frac{31}{5}\)
In simple words: Find cos theta using the Pythagorean identity, check quadrant signs, compute tan and sec, and substitute them into the given expression.

Exam Tip: Always double-check signs according to the quadrant before calculating secant and tangent values.

Question 6. Prove that :- (i) \(\frac{\cos(2\pi + \theta)\csc(2\pi + \theta)\tan\left(\frac{\pi}{2} + \theta\right)}{\sec\left(\frac{\pi}{2} + \theta\right)\cos\theta\cot(\pi + \theta)} = 1\)
Answer: Simplify each term in the expression using reduction formulas:
1. \(\cos(2\pi + \theta) = \cos\theta\)
2. \(\csc(2\pi + \theta) = \csc\theta\)
3. \(\tan\left(\frac{\pi}{2} + \theta\right) = -\cot\theta = -\frac{\cos\theta}{\sin\theta}\)
4. \(\sec\left(\frac{\pi}{2} + \theta\right) = -\csc\theta = -\frac{1}{\sin\theta}\)
5. \(\cos\theta = \cos\theta\)
6. \(\cot(\pi + \theta) = \cot\theta = \frac{\cos\theta}{\sin\theta}\)
Substitute these into the Left Hand Side:
\(\text{LHS} = \frac{\cos\theta \cdot \csc\theta \cdot (-\cot\theta)}{(-\csc\theta) \cdot \cos\theta \cdot \cot\theta}\)
Cancel the common terms \(\cos\theta\), \(\csc\theta\), and \(\cot\theta\) from numerator and denominator:
\(= \frac{-1}{-1} = 1 = \text{RHS}\)
Hence proved.
In simple words: Break down each compound angle trigonometric term using standard quadrant reduction rules and simplify.

Exam Tip: Memorize standard allied angle formulas (like \(90^\circ \pm \theta\), \(180^\circ \pm \theta\), \(360^\circ \pm \theta\)) to quickly simplify terms.

Question 6. (ii) \(\frac{\sin(180^\circ + \theta)\csc(360^\circ - \theta)\cos(90^\circ - \theta)}{\tan(90^\circ - \theta)\sec(180^\circ - \theta)\sin(-\theta)} = 1\)
Answer: Simplify each term using reduction formulas:
1. \(\sin(180^\circ + \theta) = -\sin\theta\)
2. \(\csc(360^\circ - \theta) = -\csc\theta\)
3. \(\cos(90^\circ - \theta) = \sin\theta\)
4. \(\tan(90^\circ - \theta) = \cot\theta = \frac{\cos\theta}{\sin\theta}\)
5. \(\sec(180^\circ - \theta) = -\sec\theta = -\frac{1}{\cos\theta}\)
6. \(\sin(-\theta) = -\sin\theta\)
Substitute these into the Left Hand Side:
\(\text{LHS} = \frac{(-\sin\theta)(-\csc\theta)(\sin\theta)}{(\cot\theta)(-\sec\theta)(-\sin\theta)}\)
Since \((-\sin\theta)(-\csc\theta) = (-\sin\theta)\left(-\frac{1}{\sin\theta}\right) = 1\), the numerator simplifies to:
Numerator = \(1 \cdot \sin\theta = \sin\theta\)
Denominator = \(\left(\frac{\cos\theta}{\sin\theta}\right) \cdot \left(-\frac{1}{\cos\theta}\right) \cdot (-\sin\theta) = \left(-\frac{1}{\sin\theta}\right) \cdot (-\sin\theta) = 1\)
Wait, let's re-verify the denominator carefully:
\(\text{Denominator} = \cot\theta \cdot (-\sec\theta) \cdot (-\sin\theta) = \left(\frac{\cos\theta}{\sin\theta}\right) \cdot \left(\frac{-1}{\cos\theta}\right) \cdot (-\sin\theta)\)
\(= \left(-\frac{1}{\sin\theta}\right) \cdot (-\sin\theta) = 1\)
Thus, \(\text{LHS} = \frac{-\sin^2\theta \cdot (-\frac{1}{\sin\theta}) \cdot \sin\theta}{\dots}\) Let's do it term by term:
Numerator = \((-\sin\theta) \cdot \left(-\frac{1}{\sin\theta}\right) \cdot \sin\theta = \sin\theta\)
Denominator = \(\left(\frac{\cos\theta}{\sin\theta}\right) \cdot (-\sec\theta) \cdot (-\sin\theta) = \left(\frac{\cos\theta}{\sin\theta}\right) \cdot \left(-\frac{1}{\cos\theta}\right) \cdot (-\sin\theta) = \left(-\frac{1}{\sin\theta}\right)(-\sin\theta) = 1\)
Wait, let's re-evaluate: Numerator = \((-\sin\theta)(-\csc\theta)\sin\theta = (-\sin\theta)\left(-\frac{1}{\sin\theta}\right)\sin\theta = \sin\theta\).
Denominator = \(\tan(90^\circ-\theta)\sec(180^\circ-\theta)\sin(-\theta) = \cot\theta(-\sec\theta)(-\sin\theta) = \left(\frac{\cos\theta}{\sin\theta}\right)\left(-\frac{1}{\cos\theta}\right)(-\sin\theta) = 1\).
Wait, \(\sin\theta / 1 = \sin\theta\)? Let's check original question text: RHS is 1.
Let's re-verify the exact terms: Numerator has \(\sin(180^\circ+\theta) = -\sin\theta\), \(\csc(360^\circ-\theta) = -\csc\theta\), \(\cos(90^\circ-\theta) = \sin\theta\). Product = \((-\sin\theta)(-\csc\theta)(\sin\theta) = \sin\theta\).
Denominator has \(\tan(90^\circ-\theta) = \cot\theta\), \(\sec(180^\circ-\theta) = -\sec\theta\), \(\sin(-\theta) = -\sin\theta\). Product = \(\cot\theta(-\sec\theta)(-\sin\theta) = \frac{\cos\theta}{\sin\theta} \cdot \frac{-1}{\cos\theta} \cdot (-\sin\theta) = 1\).
So \(\text{LHS} = \frac{\sin\theta}{1} = \sin\theta\)? Wait, let's check if there's a typo in the OCR text or if a term was missed. Ah, OCR says: \(\frac{\sin(180^\circ+\theta)\csc(360^\circ-\theta)\cos(90^\circ-\theta)}{\tan(90^\circ-\theta)\sec(180^\circ-\theta)\sin(-\theta)}\). If numerator has extra term or if it evaluates to 1, let's write out the standard textbook problem solution:
Let's re-verify standard identity: Usually this equals 1 if numerator has \(\sin\theta\). Let's present the clean step-by-step reduction:
\(\sin(180^\circ + \theta) = -\sin\theta\)
\(\csc(360^\circ - \theta) = -\csc\theta\)
\(\cos(90^\circ - \theta) = \sin\theta\)
\(\tan(90^\circ - \theta) = \cot\theta\)
\(\sec(180^\circ - \theta) = -\sec\theta\)
\(\sin(-\theta) = -\sin\theta\)
Numerator = \((-\sin\theta)(-\csc\theta)(\sin\theta) = \sin\theta\)
Denominator = \(\cot\theta(-\sec\theta)(-\sin\theta) = 1\)
Therefore, \(\text{LHS} = \sin\theta\). (Note: If the problem as printed intended to prove equals 1, ensure all bracketed signs match standard textbook problems).
In simple words: Substitute each trigonometric reduction formula carefully to simplify the expression.

Exam Tip: Be careful with negative signs when dealing with functions in the third and fourth quadrants.

Question 6. (iii) \(\frac{1 - \tan 69^\circ \tan 66^\circ}{\tan 69^\circ + \tan 66^\circ} = -1\)
Answer: Consider the Left Hand Side (LHS):
\(\text{LHS} = \frac{1 - \tan 69^\circ \tan 66^\circ}{\tan 69^\circ + \tan 66^\circ}\)
Notice that this resembles the reciprocal of the tangent addition formula: \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
Here, let \(A = 69^\circ\) and \(B = 66^\circ\).
Then the expression is \(\frac{1}{\tan(69^\circ + 66^\circ)} = \cot(69^\circ + 66^\circ)\)
Calculate \(69^\circ + 66^\circ\):
\(69^\circ + 66^\circ = 135^\circ\)
So the expression becomes:
\(\cot(135^\circ) = \cot(180^\circ - 45^\circ) = -\cot(45^\circ) = -1 = \text{RHS}\)
Hence proved.
In simple words: Recognize the reciprocal of the tan(A+B) formula, sum the angles to 135 degrees, and evaluate cot(135).

Exam Tip: Spotting compound angle expansion patterns immediately reduces long calculations to a single step.

Question 6. (iv) \(\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ\)
Answer: Consider the Left Hand Side (LHS):
\(\text{LHS} = \frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ}\)
Divide every term in both the numerator and the denominator by \(\cos 11^\circ\):
\(= \frac{\frac{\cos 11^\circ}{\cos 11^\circ} + \frac{\sin 11^\circ}{\cos 11^\circ}}{\frac{\cos 11^\circ}{\cos 11^\circ} - \frac{\sin 11^\circ}{\cos 11^\circ}}\)
\(= \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ}\)
We know that \(\tan(45^\circ) = 1\). Substitute \(1\) with \(\tan(45^\circ)\):
\(= \frac{\tan 45^\circ + \tan 11^\circ}{1 - \tan 45^\circ \tan 11^\circ}\)
This matches the formula for \(\tan(A + B)\) where \(A = 45^\circ\) and \(B = 11^\circ\):
\(= \tan(45^\circ + 11^\circ)\)
\(= \tan 56^\circ = \text{RHS}\)
Hence proved.
In simple words: Divide numerator and denominator by cos 11 to convert the expression into the tan(45 + 11) formula.

Exam Tip: The \(\frac{\cos\theta \pm \sin\theta}{\cos\theta \mp \sin\theta}\) form can always be simplified by dividing through by \(\cos\theta\).

Question 7. If \(\tan\alpha = \frac{m}{m + 1}\), \(\tan\beta = \frac{1}{2m + 1}\) then prove that \(\alpha + \beta = \frac{\pi}{4}\)
Answer: Consider the formula for \(\tan(\alpha + \beta)\):
\(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\)
Substitute the given values of \(\tan\alpha\) and \(\tan\beta\):
Numerator: \(\frac{m}{m + 1} + \frac{1}{2m + 1} = \frac{m(2m + 1) + 1(m + 1)}{(m + 1)(2m + 1)}\)
\(= \frac{2m^2 + m + m + 1}{(m + 1)(2m + 1)} = \frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}\)
Denominator: \(1 - \left(\frac{m}{m + 1}\right)\left(\frac{1}{2m + 1}\right) = 1 - \frac{m}{(m + 1)(2m + 1)}\)
\(= \frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)} = \frac{2m^2 + m + 2m + 1 - m}{(m + 1)(2m + 1)} = \frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}\)
Now, put numerator and denominator together in the tangent formula:
\(\tan(\alpha + \beta) = \frac{\frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}}{\frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}} = 1\)
Since \(\tan(\alpha + \beta) = 1\) and \(\alpha, \beta\) are acute, we have:
\(\alpha + \beta = \frac{\pi}{4}\)
Hence proved.
In simple words: Substitute tan alpha and tan beta into the tan(a+b) formula, simplify both numerator and denominator to get 1, and deduce the angle sum.

Exam Tip: When denominators of fractions in numerator and denominator are identical, they cancel out cleanly.

Question 8. Prove that (i) \(\tan 3A \tan 2A \tan A = \tan 3A - \tan 2A - \tan A\)
Answer: Start with the angle relation:
\(3A = 2A + A\)
Apply tangent on both sides:
\(\tan(3A) = \tan(2A + A)\)
Expand using the tangent addition formula:
\(\tan 3A = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}\)
Cross-multiply to clear the fraction:
\(\tan 3A (1 - \tan 2A \tan A) = \tan 2A + \tan A\)
\(\tan 3A - \tan 3A \tan 2A \tan A = \tan 2A + \tan A\)
Rearrange terms to isolate the product term:
\(\tan 3A \tan 2A \tan A = \tan 3A - \tan 2A - \tan A\)
Hence proved.
In simple words: Write 3A as 2A + A, apply the tan addition formula, and cross-multiply to rearrange the terms.

Exam Tip: Splitting a larger angle into the sum of two smaller angles is a standard technique for product-to-difference identities involving tangents.

Question 8. (ii) \((1 + \tan A)(1 + \tan B) = 2\) when \(A + B = \frac{\pi}{4}\)
Answer: Given \(A + B = \frac{\pi}{4}\).
Take tangent on both sides:
\(\tan(A + B) = \tan\left(\frac{\pi}{4}\right) = 1\)
Expand using the tangent addition formula:
\(\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1\)
Cross-multiply:
\(\tan A + \tan B = 1 - \tan A \tan B\)
Rearrange terms:
\(\tan A + \tan B + \tan A \tan B = 1\)
Add \(1\) to both sides of the equation:
\(1 + \tan A + \tan B + \tan A \tan B = 1 + 1\)
Factor the left-hand side by grouping:
\(1(1 + \tan A) + \tan B(1 + \tan A) = 2\)
\((1 + \tan A)(1 + \tan B) = 2\)
Hence proved.
In simple words: Apply tan to both sides of A + B = pi/4, cross-multiply, and add 1 to both sides to factor by grouping.

Exam Tip: Adding 1 to both sides of \(\tan A + \tan B + \tan A\tan B = 1\) is a classic trick for factorizing into binomial products.

Question 9. Draw the graph of (i) \(y = 3\sin x\) (ii) \(y = \csc x\) (iii) \(y = \sec x\) (iv) \(y = \sin x + \cos x\)
Answer:
(i) \(y = 3\sin x\): This is a sine wave with its amplitude stretched vertically by a factor of 3. The maximum value is \(3\) and the minimum value is \(-3\), with a period of \(2\pi\).
(ii) \(y = \csc x\): This is the cosecant graph, which is the reciprocal of \(\sin x\). It has vertical asymptotes wherever \(\sin x = 0\) (i.e., at \(x = 0, \pm\pi, \pm2\pi, \dots\)) and ranges from \((-\infty, -1] \cup [1, \infty)\).
(iii) \(y = \sec x\): This is the secant graph, which is the reciprocal of \(\cos x\). It has vertical asymptotes wherever \(\cos x = 0\) (i.e., at \(x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \dots\)) and ranges from \((-\infty, -1] \cup [1, \infty)\).
(iv) \(y = \sin x + \cos x\): Express this as a single sinusoid by rewriting it as \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x\right) = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)\). This is a sine wave with an amplitude of \(\sqrt{2}\) and a phase shift of \(-\frac{\pi}{4}\).
In simple words: Sketch the standard trigonometric curves adjusting for amplitude scaling (for 3 sin x), reciprocal asymptotes (for csc and sec), and phase/amplitude combination (for sin x + cos x).

Exam Tip: For combined trigonometric functions like \(\sin x + \cos x\), converting to \(R\sin(x + \alpha)\) makes plotting amplitude and phase shifts straightforward.

Question 10. Solve the following trigonometric equations :- (i) \(\tan\left(\frac{2}{3}\theta\right) = \sqrt{3}\)
Answer: Given equation:
\(\tan\left(\frac{2}{3}\theta\right) = \sqrt{3}\)
We know that \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\).
Using the general solution for tangent equations \(\tan x = \tan y \implies x = n\pi + y\), where \(n \in \mathbb{Z}\):
\(\frac{2}{3}\theta = n\pi + \frac{\pi}{3}\)
Multiply both sides by \(\frac{3}{2}\) to solve for \(\theta\):
\(\theta = \frac{3}{2}\left(n\pi + \frac{\pi}{3}\right)\)
\(\theta = \frac{3n\pi}{2} + \frac{\pi}{2}\)
\(\theta = \frac{3n + 1}{2}\pi\), where \(n \in \mathbb{Z}\).
In simple words: Equate the angle to the principal value plus \(n\pi\), then isolate theta.

Exam Tip: Always state that \(n\) is an integer (\(n \in \mathbb{Z}\)) when writing general solutions.

Question 10. (ii) \(7\cos^2\theta + 3\sin^2\theta = 4\)
Answer: Given equation:
\(7\cos^2\theta + 3\sin^2\theta = 4\)
Convert \(\sin^2\theta\) into \(1 - \cos^2\theta\) using the fundamental identity:
\(7\cos^2\theta + 3(1 - \cos^2\theta) = 4\)
\(7\cos^2\theta + 3 - 3\cos^2\theta = 4\)
\(4\cos^2\theta + 3 = 4\)
\(4\cos^2\theta = 1\)
\(\cos^2\theta = \frac{1}{4}\)
\(\cos^2\theta = \left(\frac{1}{2}\right)^2 = \cos^2\left(\frac{\pi}{3}\right)\)
Using the general solution for \(\cos^2\theta = \cos^2\alpha\):
\(\theta = n\pi \pm \frac{\pi}{3}\), where \(n \in \mathbb{Z}\).
In simple words: Convert all sine terms to cosine using \(\sin^2\theta = 1 - \cos^2\theta\), simplify, and apply the general solution formula.

Exam Tip: Expressing equations entirely in terms of sine or entirely in terms of cosine is the key to solving homogeneous trigonometric equations.

Question 11. If \(\tan A = x\tan B\) then prove that \(\frac{\sin(A - B)}{\sin(A + B)} = \frac{x - 1}{x + 1}\)
Answer: Given \(\tan A = x\tan B\), which can be written as:
\(\frac{\tan A}{\tan B} = x\)
Express in terms of sine and cosine:
\(\frac{\frac{\sin A}{\cos A}}{\frac{\sin B}{\cos B}} = x \implies \frac{\sin A \cos B}{\cos A \sin B} = x\)
Apply Componendo and Dividendo on \(\frac{\sin A \cos B}{\cos A \sin B} = \frac{x}{1}\):
\(\frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B} = \frac{x + 1}{x - 1}\)
Notice that the numerator and denominator are standard sine expansion formulas:
Numerator = \(\sin(A + B)\)
Denominator = \(\sin(A - B)\)
So we have:
\(\frac{\sin(A + B)}{\sin(A - B)} = \frac{x + 1}{x - 1}\)
Take the reciprocal on both sides to get the desired result:
\(\frac{\sin(A - B)}{\sin(A + B)} = \frac{x - 1}{x + 1}\)
Hence proved.
In simple words: Convert tans to sines and cosines, write the ratio as x/1, and apply componendo and dividendo.

Exam Tip: Componendo and dividendo (\(\frac{a+b}{a-b} = \frac{c+d}{c-d}\)) is a powerful tool when proving fraction relationships in trigonometry.

Question 12. If \(\sin A = \frac{3}{5}, 0 < A < \frac{\pi}{2}\), \(\cos B = -\frac{12}{13}, \pi < B < \frac{3\pi}{2}\), find \(\sin(A + B)\)
Answer: We need to find \(\sin(A + B)\) using the expansion formula:
\(\sin(A + B) = \sin A \cos B + \cos A \sin B\)
Given \(\sin A = \frac{3}{5}\) and \(0 < A < \frac{\pi}{2}\) (First quadrant, where all ratios are positive):
\(\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}\)
Given \(\cos B = -\frac{12}{13}\) and \(\pi < B < \frac{3\pi}{2}\) (Third quadrant, where sine is negative):
\(\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \left(-\frac{12}{13}\right)^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}\)
Now, substitute all values into the formula for \(\sin(A + B)\):
\(\sin(A + B) = \left(\frac{3}{5}\right)\left(-\frac{12}{13}\right) + \left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)\)
\(= -\frac{36}{65} - \frac{20}{65}\)
\(= \frac{-36 - 20}{65} = -\frac{56}{65}\)
In simple words: Find cos A and sin B using Pythagorean identities and correct quadrant signs, then plug them into the sin(A+B) formula.

Exam Tip: Check quadrant signs carefully: in the third quadrant, cosine and sine are both negative.

Question 13. Prove that (i) \(\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16}\)
Answer: Consider the Left Hand Side (LHS):
\(\text{LHS} = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ\)
We know that \(\cos 60^\circ = \frac{1}{2}\). Substitute this value:
\(= \frac{1}{2} \cos 20^\circ \cos 40^\circ \cos 80^\circ\)
Multiply and divide the expression by \(2\sin 20^\circ\):
\(= \frac{1}{2 \cdot 2\sin 20^\circ} [2\sin 20^\circ \cos 20^\circ] \cos 40^\circ \cos 80^\circ\)
Using the double angle formula \(2\sin\theta\cos\theta = \sin(2\theta)\), \(2\sin 20^\circ \cos 20^\circ = \sin 40^\circ\):
\(= \frac{1}{4\sin 20^\circ} [\sin 40^\circ \cos 40^\circ] \cos 80^\circ\)
Multiply and divide by 2 inside the bracket:
\(= \frac{1}{4 \cdot 2\sin 20^\circ} [2\sin 40^\circ \cos 40^\circ] \cos 80^\circ\)
Since \(2\sin 40^\circ \cos 40^\circ = \sin 80^\circ\):
\(= \frac{1}{8\sin 20^\circ} [\sin 80^\circ] \cos 80^\circ\)
Again, multiply and divide by 2:
\(= \frac{1}{8 \cdot 2\sin 20^\circ} [2\sin 80^\circ \cos 80^\circ]\)
Since \(2\sin 80^\circ \cos 80^\circ = \sin(160^\circ)\):
\(= \frac{1}{16\sin 20^\circ} \sin(160^\circ)\)
Since \(\sin(160^\circ) = \sin(180^\circ - 20^\circ) = \sin 20^\circ\):
\(= \frac{1}{16\sin 20^\circ} (\sin 20^\circ) = \frac{1}{16} = \text{RHS}\)
Hence proved.
In simple words: Substitute cos 60 = 1/2, then repeatedly multiply and divide by 2 and use the 2 sin theta cos theta formula.

Exam Tip: Multiplying and dividing by \(2\sin\theta\) (where \(\theta\) is the smallest angle) is a standard technique for product series of cosines.

Question 13. (ii) \(\cos 5A = 16\cos^5 A - 20\cos^3 A + 5\cos A\)
Answer: Express \(5A\) as \(3A + 2A\):
\(\cos(5A) = \cos(3A + 2A)\)
Apply the cosine addition formula: \(\cos(X + Y) = \cos X \cos Y - \sin X \sin Y\)
\(\cos(5A) = \cos 3A \cos 2A - \sin 3A \sin 2A\)
Expand each term using multiple angle formulas in terms of cosine and sine:
\(\cos 3A = 4\cos^3 A - 3\cos A\)
\(\cos 2A = 2\cos^2 A - 1\)
\(\sin 3A = 3\sin A - 4\sin^3 A = \sin A (3 - 4\sin^2 A)\)
\(\sin 2A = 2\sin A \cos A\)
Substitute these expressions:
\(\cos 5A = (4\cos^3 A - 3\cos A)(2\cos^2 A - 1) - (\sin A (3 - 4\sin^2 A))(2\sin A \cos A)\)
\(= (4\cos^3 A - 3\cos A)(2\cos^2 A - 1) - 2\sin^2 A \cos A (3 - 4\sin^2 A)\)
Substitute \(\sin^2 A = 1 - \cos^2 A\) into the second term:
\(= (4\cos^3 A - 3\cos A)(2\cos^2 A - 1) - 2(1 - \cos^2 A)\cos A [3 - 4(1 - \cos^2 A)]\)
\(= (8\cos^5 A - 4\cos^3 A - 6\cos^3 A + 3\cos A) - 2\cos A (1 - \cos^2 A)(3 - 4 + 4\cos^2 A)\)
\(= (8\cos^5 A - 10\cos^3 A + 3\cos A) - 2\cos A (1 - \cos^2 A)(4\cos^2 A - 1)\)
Expand the second product:
\(2\cos A (4\cos^2 A - 1 - 4\cos^4 A + \cos^2 A) = 2\cos A (5\cos^2 A - 1 - 4\cos^4 A)\)
\(= 10\cos^3 A - 2\cos A - 8\cos^5 A\)
Now substitute this back:
\(\cos 5A = 8\cos^5 A - 10\cos^3 A + 3\cos A - (10\cos^3 A - 2\cos A - 8\cos^5 A)\)
\(= 8\cos^5 A - 10\cos^3 A + 3\cos A - 10\cos^3 A + 2\cos A + 8\cos^5 A\)
Combine like terms:
\(= (8 + 8)\cos^5 A + (-10 - 10)\cos^3 A + (3 + 2)\cos A\)
\(= 16\cos^5 A - 20\cos^3 A + 5\cos A = \text{RHS}\)
Hence proved.
In simple words: Write 5A as 3A + 2A, expand using cosine addition, substitute multiple angle formulas, and convert remaining sines to cosines.

Exam Tip: Keep careful track of signs and brackets when expanding multiple angle trigonometric identities.

Question 14. Find the value of (i) \(\cos 15^\circ\) (ii) \(\sin 75^\circ\) (iii) \(\tan 75^\circ\)
Answer:
(i) \(\cos 15^\circ\):
Write \(15^\circ = 45^\circ - 30^\circ\)
\(\cos 15^\circ = \cos(45^\circ - 30^\circ)\)
Use the cosine difference formula: \(\cos(A - B) = \cos A \cos B + \sin A \sin B\)
\(= \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ\)
\(= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)
\(= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}\)
Rationalize the denominator:
\(= \frac{\sqrt{6} + \sqrt{2}}{4}\)

(ii) \(\sin 75^\circ\):
Write \(75^\circ = 45^\circ + 30^\circ\)
\(\sin 75^\circ = \sin(45^\circ + 30^\circ)\)
Use the sine addition formula: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)
\(= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ\)
\(= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)
\(= \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

(iii) \(\tan 75^\circ\):
Write \(75^\circ = 45^\circ + 30^\circ\)
\(\tan 75^\circ = \tan(45^\circ + 30^\circ)\)
Use the tangent addition formula: \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
\(= \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}\)
\(= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}\)
\(= \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\)
Rationalize the denominator:
\(= \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}\)
In simple words: Express angles as sums or differences of standard angles (30, 45, 60 degrees) and apply compound angle formulas.

Exam Tip: Standard values like \(\sin 15^\circ\), \(\cos 15^\circ\), and \(\tan 75^\circ\) frequently appear and can be memorized as standard results.

Question 15. Find the general solution of the following trigonometric equations :- (i) \(\cot^2\theta + \frac{3}{\sin\theta} + 3 = 0\)
Answer: Given equation:
\(\cot^2\theta + \frac{3}{\sin\theta} + 3 = 0\)
Rewrite \(\cot^2\theta\) as \(\csc^2\theta - 1\):
\((\csc^2\theta - 1) + \frac{3}{\sin\theta} + 3 = 0\)
\(\csc^2\theta + \frac{3}{\sin\theta} + 2 = 0\)
Since \(\csc\theta = \frac{1}{\sin\theta}\), let \(x = \frac{1}{\sin\theta}\):
\(x^2 + 3x + 2 = 0\)
Factor the quadratic equation:
\((x + 1)(x + 2) = 0\)
So, \(x = -1\) or \(x = -2\)
Case 1: \(\csc\theta = -1 \implies \sin\theta = -1\)
The general solution for \(\sin\theta = -1\) is:
\(\theta = n\pi + (-1)^n \left(-\frac{\pi}{2}\right)\), where \(n \in \mathbb{Z}\).
Case 2: \(\csc\theta = -2 \implies \sin\theta = -\frac{1}{2} = -\sin\left(\frac{\pi}{6}\right) = \sin\left(\pi + \frac{\pi}{6}\right) = \sin\left(\frac{7\pi}{6}\right)\)
The general solution for \(\sin\theta = \sin\alpha\) is \(\theta = n\pi + (-1)^n \alpha\):
\(\theta = n\pi + (-1)^n \left(-\frac{\pi}{6}\right)\) or similar standard form.
In simple words: Convert cot squared into cosec squared, substitute x = 1/sin theta to form a quadratic equation, solve for x, and find general solutions for sine.

Exam Tip: Converting mixed trigonometric functions into a single function using identities is standard for solving trigonometric equations.

Question 15. (ii) \(\tan\theta + \tan 2\theta + \tan\theta \tan 2\theta = 1\)
Answer: Given equation:
\(\tan\theta + \tan 2\theta + \tan\theta \tan 2\theta = 1\)
Rearrange terms:
\(\tan\theta + \tan 2\theta = 1 - \tan\theta \tan 2\theta\)
Divide both sides by \(1 - \tan\theta \tan 2\theta\):
\(\frac{\tan\theta + \tan 2\theta}{1 - \tan\theta \tan 2\theta} = 1\)
The left side is the formula for \(\tan(\theta + 2\theta)\):
\(\tan(3\theta) = 1\)
We know that \(\tan\left(\frac{\pi}{4}\right) = 1\).
Using the general solution for tangent equations \(\tan X = \tan Y \implies X = n\pi + Y\):
\(3\theta = n\pi + \frac{\pi}{4}\)
\(\theta = \frac{n\pi}{3} + \frac{\pi}{12}\), where \(n \in \mathbb{Z}\).
In simple words: Rearrange the equation into the tan addition formula form, simplify to tan(3theta) = 1, and find the general solution.

Exam Tip: Recognizing standard expansion forms in reverse order is essential for solving advanced trigonometric equations.

Question 15. (iii) \(\tan\theta + \tan 2\theta + 3\tan\theta \tan 2\theta = 3\)
Answer: Given equation:
\(\tan\theta + \tan 2\theta + 3\tan\theta \tan 2\theta = 3\)
Rewrite as:
\(\tan\theta + \tan 2\theta = 3(1 - \tan\theta \tan 2\theta)\)
Divide both sides by \(1 - \tan\theta \tan 2\theta\):
\(\frac{\tan\theta + \tan 2\theta}{1 - \tan\theta \tan 2\theta} = 3\)
\(\tan(3\theta) = 3\)
Using the general solution for tangent equations \(\tan X = 3\):
\(3\theta = n\pi + \tan^{-1}(3)\)
\(\theta = \frac{n\pi}{3} + \frac{1}{3}\tan^{-1}(3)\), where \(n \in \mathbb{Z}\).
In simple words: Divide to form tan(3theta) = 3, then use inverse tangent for the general solution.

Exam Tip: When the value is not a standard trigonometric angle, express the principal solution using inverse trigonometric notation \(\tan^{-1}(k)\).

Question 15. (iv) \(\sec^2 2x = 1 - \tan 2x\)
Answer: Given equation:
\(\sec^2 2x = 1 - \tan 2x\)
Use the fundamental identity \(\sec^2\theta = 1 + \tan^2\theta\):
\(1 + \tan^2 2x = 1 - \tan 2x\)
Cancel 1 from both sides:
\(\tan^2 2x = -\tan 2x\)
\(\tan^2 2x + \tan 2x = 0\)
Factor out \(\tan 2x\):
\(\tan 2x (\tan 2x + 1) = 0\)
This gives two cases:
Case 1: \(\tan 2x = 0\)
\(2x = n\pi \implies x = \frac{n\pi}{2}\), where \(n \in \mathbb{Z}\).
Case 2: \(\tan 2x + 1 = 0 \implies \tan 2x = -1 = -\tan\left(\frac{\pi}{4}\right) = \tan\left(\pi - \frac{\pi}{4}\right) = \tan\left(\frac{3\pi}{4}\right)\)
\(2x = n\pi + \frac{3\pi}{4} \implies x = \frac{n\pi}{2} + \frac{3\pi}{8}\), where \(n \in \mathbb{Z}\).
In simple words: Replace sec squared with 1 + tan squared, factor the resulting quadratic in tan, and solve both cases for general solutions.

Exam Tip: Factoring equations with trigonometric terms is similar to algebraic factoring; never cancel a variable factor unless it cannot be zero.

Question 15. (v) \(\tan\theta + \tan\left(\theta + \frac{\pi}{3}\right) + \tan\left(\theta + \frac{2\pi}{3}\right) = 3\)
Answer: We can use the triple angle identity for tangent: \(\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\), which can be factored as the product of tangents of three angles separated by \(\frac{\pi}{3}\).
Using the standard identity for the sum of tangents: \(\tan\theta + \tan\left(\theta + \frac{\pi}{3}\right) + \tan\left(\theta + \frac{2\pi}{3}\right) = 3\tan 3\theta\)
Substitute this into the given equation:
\(3\tan 3\theta = 3\)
\(\tan 3\theta = 1\)
\(\tan 3\theta = \tan\left(\frac{\pi}{4}\right)\)
General solution:
\(3\theta = n\pi + \frac{\pi}{4}\)
\(\theta = \frac{n\pi}{3} + \frac{\pi}{12}\), where \(n \in \mathbb{Z}\).
In simple words: Use the standard triple angle relation for the sum of tangents, simplify to tan 3theta = 1, and find the general solution.

Exam Tip: Memorizing the identity \(\tan\theta + \tan\left(\theta + \frac{\pi}{3}\right) + \tan\left(\theta + \frac{2\pi}{3}\right) = 3\tan 3\theta\) saves significant time.

Question 16. In any triangle ABC, if \(a = 16, b = 12, c = 25\) find (i) \(\cos A, \cos B, \cos C\)
Answer: Given side lengths: \(a = 16, b = 12, c = 25\).
Use the Law of Cosines to find each cosine value:
1. For \(\cos A\):
\(\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{12^2 + 25^2 - 16^2}{2(12)(25)}\)
\(= \frac{144 + 625 - 256}{600} = \frac{513}{600} = \frac{171}{200} = 0.855\)
2. For \(\cos B\):
\(\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{16^2 + 25^2 - 12^2}{2(16)(25)}\)
\(= \frac{256 + 625 - 144}{800} = \frac{737}{800} = 0.92125\)
3. For \(\cos C\):
\(\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{16^2 + 12^2 - 25^2}{2(16)(12)}\)
\(= \frac{256 + 144 - 625}{384} = \frac{-225}{384} = -\frac{75}{128} = -0.5859375\)
In simple words: Apply the Law of Cosines formula for each angle using the given side lengths.

Exam Tip: The Law of Cosines is essential when all three sides of a triangle are known and angles need to be found.

Question 16. (ii) \(\sin A, \sin B, \sin C\)
Answer: Using the relation \(\sin\theta = \sqrt{1 - \cos^2\theta}\) for each angle (since angles of a triangle are between \(0^\circ\) and \(180^\circ\), sine is always positive):
1. For \(\sin A\):
\(\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{513}{600}\right)^2} = \sqrt{1 - \frac{263169}{360000}} = \sqrt{\frac{96831}{360000}} \approx \frac{311.18}{600} \approx 0.5186\)
2. For \(\sin B\):
\(\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{737}{800}\right)^2} = \sqrt{1 - \frac{543169}{640000}} = \sqrt{\frac{96831}{640000}} \approx \frac{311.18}{800} \approx 0.3890\)
3. For \(\sin C\):
\(\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \left(-\frac{75}{128}\right)^2} = \sqrt{1 - \frac{5625}{16384}} = \sqrt{\frac{10759}{16384}} \approx \frac{103.72}{128} \approx 0.8103\)
In simple words: Compute sine from cosine using the fundamental identity \(\sin\theta = \sqrt{1 - \cos^2\theta}\).

Exam Tip: Alternatively, you can use the Sine Rule \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) once the circumradius is known.

Question 17. For any triangle ABC prove that (i) \(\frac{\sin(B - C)}{\sin(B + C)} = \frac{b^2 - c^2}{a^2}\)
Answer: In any triangle \(ABC\), \(A + B + C = \pi \implies B + C = \pi - A\).
Therefore, \(\sin(B + C) = \sin(\pi - A) = \sin A\).
Substitute this into the denominator of the Left Hand Side:
\(\text{LHS} = \frac{\sin(B - C)}{\sin A}\)
Using the Sine Rule, \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\), so \(\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}\):
\(\text{LHS} = \frac{\sin B \cos C - \cos B \sin C}{\sin A}\)
Substitute sine values using the Sine Rule and cosine values using the Law of Cosines:
Alternatively, using \(\frac{\sin(B - C)}{\sin A} = \frac{\sin(B - C)}{\sin(B + C)}\) and converting sides using the Sine Rule \(b = k\sin B, c = k\sin C, a = k\sin A\):
\(\frac{b^2 - c^2}{a^2} = \frac{k^2\sin^2 B - k^2\sin^2 C}{k^2\sin^2 A} = \frac{\sin^2 B - \sin^2 C}{\sin^2 A}\)
Using the identity \(\sin^2 B - \sin^2 C = \sin(B + C)\sin(B - C)\):
\(= \frac{\sin(B + C)\sin(B - C)}{\sin^2 A}\)
Since \(\sin(B + C) = \sin(\pi - A) = \sin A\):
\(= \frac{\sin A \sin(B - C)}{\sin^2 A} = \frac{\sin(B - C)}{\sin A} = \frac{\sin(B - C)}{\sin(B + C)} = \text{RHS}\)
Hence proved.
In simple words: Replace side lengths with sine equivalents using the Sine Rule, expand the numerator using angle difference identities, and simplify.

Exam Tip: The Sine Rule substitution \(a = k\sin A\) is extremely effective for converting geometric side ratios into angle relations.

Question 17. (ii) \(\frac{b^2 - c^2}{a^2}\sin 2A + \frac{c^2 - a^2}{b^2}\sin 2B + \frac{a^2 - b^2}{c^2}\sin 2C = 0\)
Answer: Using the Sine Rule, let \(a = k\sin A, b = k\sin B, c = k\sin C\).
Substitute these into the first term: \(\frac{b^2 - c^2}{a^2}\sin 2A\)
\(= \frac{k^2\sin^2 B - k^2\sin^2 C}{k^2\sin^2 A} (2\sin A \cos A) = \frac{\sin^2 B - \sin^2 C}{\sin^2 A} (2\sin A \cos A)\)
Using \(\sin^2 B - \sin^2 C = \sin(B + C)\sin(B - C) = \sin A \sin(B - C)\):
\(= \frac{\sin A \sin(B - C)}{\sin^2 A} (2\sin A \cos A) = 2\sin(B - C)\cos A\)
Since \(A = \pi - (B + C)\), we have \(\cos A = -\cos(B + C)\).
So the term becomes: \(-2\sin(B - C)\cos(B + C) = -[\sin 2B - \sin 2C]\).
Similarly, the second term evaluates to \(-[\sin 2C - \sin 2A]\) and the third term to \(-[\sin 2A - \sin 2B]\).
Summing all three terms:
\(-(\sin 2B - \sin 2C) - (\sin 2C - \sin 2A) - (\sin 2A - \sin 2B) = 0 = \text{RHS}\)
Hence proved.
In simple words: Convert side lengths using the Sine Rule, simplify each term into sine difference products, and sum them to zero.

Exam Tip: Cyclic symmetry in triangle identities often means that once you simplify the first term, the others follow by cyclic permutation of angles.

Question 17. (iii) \(a(\cos C - \cos B) = 2(b - c)\cos^2\left(\frac{A}{2}\right)\)
Answer: Consider the Left Hand Side (LHS):
\(\text{LHS} = a(\cos C - \cos B)\)
Using the Sine Rule \(a = k\sin A\) and the formula for difference of cosines \(\cos C - \cos B = 2\sin\left(\frac{B + C}{2}\right)\sin\left(\frac{B - C}{2}\right)\):
\(= k\sin A \cdot 2\sin\left(\frac{B + C}{2}\right)\sin\left(\frac{B - C}{2}\right)\)
Since \(A + B + C = \pi\), we have \(\frac{B + C}{2} = \frac{\pi}{2} - \frac{A}{2}\), so \(\sin\left(\frac{B + C}{2}\right) = \cos\left(\frac{A}{2}\right)\).
Also, \(\sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\).
Substitute these:
\(= k \cdot \left[2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\right] \cdot 2\cos\left(\frac{A}{2}\right) \sin\left(\frac{B - C}{2}\right)\)
\(= 4k \sin\left(\frac{A}{2}\right) \cos^2\left(\frac{A}{2}\right) \sin\left(\frac{B - C}{2}\right)\)
Using Sine Rule differences for \(b - c\): \(b - c = k(\sin B - \sin C) = 2k\sin\left(\frac{B - C}{2}\right)\cos\left(\frac{B + C}{2}\right) = 2k\sin\left(\frac{B - C}{2}\right)\sin\left(\frac{A}{2}\right)\).
Substituting this into our expression:
\(= 2 \cdot [2k\sin\left(\frac{B - C}{2}\right)\sin\left(\frac{A}{2}\right)] \cos^2\left(\frac{A}{2}\right)\)
\(= 2(b - c)\cos^2\left(\frac{A}{2}\right) = \text{RHS}\)
Hence proved.
In simple words: Use the Sine Rule and cosine difference formulas to convert both sides into matching half-angle expressions.

Exam Tip: Half-angle substitutions like \(\sin\left(\frac{B+C}{2}\right) = \cos\left(\frac{A}{2}\right)\) are standard in triangle identity proofs.