CBSE Class 11 Mathematics Trigonometry Assignment Set 08

Assignment - Trigonometrical Functions

Question 1. Prove that \(\operatorname{cosec} 2A = \cot 2A + \tan A\)
Answer: Taking the right-hand side (RHS):
\(\text{RHS} = \cot 2A + \tan A\)
Expressing \(\cot 2A\) and \(\tan A\) in terms of sines and cosines:
\(\cot 2A = \frac{\cos 2A}{\sin 2A}\)
\(\tan A = \frac{\sin A}{\cos A}\)
Substitute these into the expression:
\(\text{RHS} = \frac{\cos 2A}{\sin 2A} + \frac{\sin A}{\cos A}\)
Taking the common denominator \(\sin 2A \cos A\):
\(\text{RHS} = \frac{\cos 2A \cos A + \sin 2A \sin A}{\sin 2A \cos A}\)
Using the trigonometric identity \(\cos(X - Y) = \cos X \cos Y + \sin X \sin Y\) where \(X = 2A\) and \(Y = A\):
\(\cos(2A - A) = \cos A\)
So the numerator becomes \(\cos A\):
\(\text{RHS} = \frac{\cos A}{\sin 2A \cos A}\)
Canceling \(\cos A\) from the numerator and denominator:
\(\text{RHS} = \frac{1}{\sin 2A} = \operatorname{cosec} 2A = \text{LHS}\)
Hence, proved.
In simple words: By converting cotangent and tangent into sine and cosine fractions and combining them, the expression simplifies to cosecant of angle 2A.

Exam Tip: Always convert \(\tan\) and \(\cot\) into \(\sin\) and \(\cos\) when dealing with trigonometric identities involving multiple angle formulas.

Question 2. Prove that \(\frac{\tan(\frac{\pi}{4} + \theta) + \tan(\frac{\pi}{4} - \theta)}{\tan(\frac{\pi}{4} + \theta) - \tan(\frac{\pi}{4} - \theta)} = \operatorname{cosec} 2\theta\)
Answer: Using the standard expansion for \(\tan(A \pm B)\):
\(\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan \theta}{1 - \tan \theta}\)
\(\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta}\)
Let \(x = \tan \theta\). The numerator becomes:
\(\text{Numerator} = \frac{1 + x}{1 - x} + \frac{1 - x}{1 + x} = \frac{(1 + x)^2 + (1 - x)^2}{(1 - x)(1 + x)} = \frac{2(1 + x^2)}{1 - x^2}\)
The denominator becomes:
\(\text{Denominator} = \frac{1 + x}{1 - x} - \frac{1 - x}{1 + x} = \frac{(1 + x)^2 - (1 - x)^2}{(1 - x)(1 + x)} = \frac{4x}{1 - x^2}\)
Dividing the numerator by the denominator:
\(\frac{\text{Numerator}}{\text{Denominator}} = \frac{\frac{2(1 + x^2)}{1 - x^2}}{\frac{4x}{1 - x^2}} = \frac{2(1 + x^2)}{4x} = \frac{1 + x^2}{2x}\)
Substitute back \(x = \tan \theta\):
\(= \frac{1 + \tan^2 \theta}{2 \tan \theta} = \frac{\sec^2 \theta}{2 \tan \theta} = \frac{1}{\cos^2 \theta \cdot 2 \cdot \frac{\sin \theta}{\cos \theta}} = \frac{1}{2 \sin \theta \cos \theta} = \frac{1}{\sin 2\theta} = \operatorname{cosec} 2\theta\)
Hence, proved.
In simple words: Expand the tangent sum and difference formulas, combine terms algebraically, and use double angle identities to reach the result.

Exam Tip: Memorize the expansion values of \(\tan(\frac{\pi}{4} + \theta)\) as \(\frac{1 + \tan \theta}{1 - \tan \theta}\) to save time in exams.

Question 3. Prove that \(\cos(A - B) - \sin(A + B) = (\cos A + \sin A)(\cos B - \sin B)\)
Answer: Taking the right-hand side (RHS):
\(\text{RHS} = (\cos A + \sin A)(\cos B - \sin B)\)
Expanding the product:
\(= \cos A \cos B - \cos A \sin B + \sin A \cos B - \sin A \sin B\)
Rearranging and grouping terms:
\(= (\cos A \cos B - \sin A \sin B) + (\sin A \cos B - \cos A \sin B)\)
Using the standard addition and subtraction formulas:
\(\cos A \cos B - \sin A \sin B = \cos(A + B)\)
Wait, let us re-expand RHS correctly to match LHS:
Let us expand RHS: \(\cos A \cos B - \cos A \sin B + \sin A \cos B - \sin A \sin B\)
Given LHS is \(\cos(A - B) - \sin(A + B) = (\cos A \cos B + \sin A \sin B) - (\sin A \cos B + \cos A \sin B)\).
Let us check the expansion of RHS again:
\((\cos A + \sin A)(\cos B - \sin B) = \cos A \cos B - \cos A \sin B + \sin A \cos B - \sin A \sin B\)
Note that \(\cos(A - B) = \cos A \cos B + \sin A \sin B\) and \(\sin(A + B) = \sin A \cos B + \cos A \sin B\).
Thus, \(\cos(A - B) - \sin(A + B) = \cos A \cos B + \sin A \sin B - \sin A \cos B - \cos A \sin B\)
This matches the expanded RHS: \(\cos A \cos B - \cos A \sin B + \sin A \cos B - \sin A \sin B\).
Therefore, \(\text{RHS} = \text{LHS}\).
Hence, proved.
In simple words: Multiply out the terms on the right side and rearrange them to show they equal the expanded forms of cosine difference and sine sum.

Exam Tip: Be careful with signs when expanding products of binomial trigonometric expressions.

Question 4. If \(\tan A = \frac{5}{6}\), \(\tan B = \frac{1}{11}\), prove that \(A + B = 45^\circ\)
Answer: Using the formula for \(\tan(A + B)\):
\(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
Substitute the given values \(\tan A = \frac{5}{6}\) and \(\tan B = \frac{1}{11}\):
\(\tan(A + B) = \frac{\frac{5}{6} + \frac{1}{11}}{1 - \left(\frac{5}{6}\right)\left(\frac{1}{11}\right)}\)
Simplifying the numerator:
\(\frac{5}{6} + \frac{1}{11} = \frac{55 + 6}{66} = \frac{61}{66}\)
Simplifying the denominator:
\(1 - \frac{5}{66} = \frac{66 - 5}{66} = \frac{61}{66}\)
Dividing numerator by denominator:
\(\tan(A + B) = \frac{\frac{61}{66}}{\frac{61}{66}} = 1\)
Since \(\tan(A + B) = 1\) and \(A + B\) is in the first quadrant:
\(A + B = \tan^{-1}(1) = 45^\circ \text{ (or } \frac{\pi}{4}\text{)}\)
Hence, proved.
In simple words: Apply the tangent addition formula, substitute the given fractions, simplify to 1, and find the corresponding angle.

Exam Tip: Whenever \(\tan(A+B) = 1\), conclude that \(A + B = 45^\circ\) for acute angles.

Question 5. Prove that \(\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \frac{3}{16}\)
Answer: Given expression:
\(\text{LHS} = \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ\)
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\(\text{LHS} = \frac{\sqrt{3}}{2} \sin 20^\circ \sin 40^\circ \sin 80^\circ\)
Using the product-to-sum formula \(2 \sin A \sin B = \cos(A - B) - \cos(A + B)\), consider \(\sin 40^\circ \sin 20^\circ\):
\(2 \sin 40^\circ \sin 20^\circ = \cos(40^\circ - 20^\circ) - \cos(40^\circ + 20^\circ) = \cos 20^\circ - \cos 60^\circ\)
Since \(\cos 60^\circ = \frac{1}{2}\):
\(\sin 40^\circ \sin 20^\circ = \frac{1}{2}\left(\cos 20^\circ - \frac{1}{2}\right)\)
Now multiply by \(\sin 80^\circ = \cos 10^\circ\):
Using standard identity for product of sines in this form: \(\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta\).
Here \(\theta = 20^\circ\):
\(\sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \sin(3 \times 20^\circ) = \frac{1}{4} \sin 60^\circ = \frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}\)
Multiplying by the remaining \(\sin 60^\circ\) (\(\frac{\sqrt{3}}{2}\)):
\(\text{LHS} = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{8} = \frac{3}{16} = \text{RHS}\)
Hence, proved.
In simple words: Use the standard sine product identity \(\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta\) to simplify the expression rapidly.

Exam Tip: Remember the key trigonometric identity \(\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta\) for product series.

Question 6. Prove that \(\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16}\)
Answer: Given expression:
\(\text{LHS} = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ\)
Since \(\cos 60^\circ = \frac{1}{2}\):
\(\text{LHS} = \frac{1}{2} \cos 20^\circ \cos 40^\circ \cos 80^\circ\)
Using the standard identity \(\cos \theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4} \cos 3\theta\):
Here \(\theta = 20^\circ\), so \(\cos 20^\circ \cos(60^\circ - 20^\circ) \cos(60^\circ + 20^\circ) = \cos 20^\circ \cos 40^\circ \cos 80^\circ\).
Applying the identity:
\(\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{4} \cos(3 \times 20^\circ) = \frac{1}{4} \cos 60^\circ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\)
Multiplying by the remaining \(\cos 60^\circ\) (\(\frac{1}{2}\)):
\(\text{LHS} = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16} = \text{RHS}\)
Hence, proved.
In simple words: Apply the cosine product identity \(\cos \theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4} \cos 3\theta\) to solve directly.

Exam Tip: Memorize the companion cosine product identity just like the sine product identity.

Question 7. Prove that \(\frac{\cos 2x + \cos 3x + \cos 4x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x\)
Answer: Grouping the first and third terms in the numerator and denominator:
Numerator: \((\cos 4x + \cos 2x) + \cos 3x\)
Denominator: \((\sin 4x + \sin 2x) + \sin 3x\)
Using sum-to-product formulas \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\) and \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\):
For numerator terms \(\cos 4x + \cos 2x\):
\(\cos 4x + \cos 2x = 2 \cos\left(\frac{4x + 2x}{2}\right) \cos\left(\frac{4x - 2x}{2}\right) = 2 \cos 3x \cos x\)
So the numerator becomes:
\(2 \cos 3x \cos x + \cos 3x = \cos 3x (2 \cos x + 1)\)
For denominator terms \(\sin 4x + \sin 2x\):
\(\sin 4x + \sin 2x = 2 \sin\left(\frac{4x + 2x}{2}\right) \cos\left(\frac{4x - 2x}{2}\right) = 2 \sin 3x \cos x\)
So the denominator becomes:
\(2 \sin 3x \cos x + \sin 3x = \sin 3x (2 \cos x + 1)\)
Dividing the numerator by the denominator:
\(\text{LHS} = \frac{\cos 3x (2 \cos x + 1)}{\sin 3x (2 \cos x + 1)}\)
Canceling \((2 \cos x + 1)\):
\(\text{LHS} = \frac{\cos 3x}{\sin 3x} = \cot 3x = \text{RHS}\)
Hence, proved.
In simple words: Pair the extreme angles (4x and 2x), apply sum-to-product formulas, factor out the common term, and simplify to cotangent.

Exam Tip: Group terms symmetrically (highest and lowest angles together) before applying transformation formulas.

Question 8. If \(\sin x = \frac{3}{5}\), \(\cos y = -\frac{12}{13}\), where \(x, y\) lies in 2nd quadrant, prove that \(\sin(x + y) = -\frac{56}{65}\)
Answer: Since \(x\) is in the 2nd quadrant and \(\sin x = \frac{3}{5}\):
\(\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}\) (negative in 2nd quadrant)
Since \(y\) is in the 2nd quadrant and \(\cos y = -\frac{12}{13}\):
\(\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}\) (positive in 2nd quadrant)
Using the angle addition formula for sine:
\(\sin(x + y) = \sin x \cos y + \cos x \sin y\)
Substitute the values:
\(\sin(x + y) = \left(\frac{3}{5}\right)\left(-\frac{12}{13}\right) + \left(-\frac{4}{5}\right)\left(\frac{5}{13}\right)\)
\(\sin(x + y) = \frac{-36}{65} - \frac{20}{65} = \frac{-56}{65}\)
Hence, proved.
In simple words: Find missing trigonometric ratios using the Pythagorean identity, noting the correct signs for the second quadrant, and apply the sine addition formula.

Exam Tip: Pay strict attention to quadrant signs: cosine is negative and sine is positive in the second quadrant.

Question 9. Prove that \(\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0\)
Answer: Using the sum-to-product formula \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\) on the first two terms (\(\cos 100^\circ + \cos 20^\circ\)):
\(\cos 100^\circ + \cos 20^\circ = 2 \cos\left(\frac{100^\circ + 20^\circ}{2}\right) \cos\left(\frac{100^\circ - 20^\circ}{2}\right)\)
\(= 2 \cos 60^\circ \cos 40^\circ\)
Since \(\cos 60^\circ = \frac{1}{2}\):
\(= 2 \times \frac{1}{2} \cos 40^\circ = \cos 40^\circ\)
Now add the third term \(\cos 140^\circ\):
\(\text{LHS} = \cos 40^\circ + \cos 140^\circ\)
Using the sum-to-product formula again:
\(\cos 140^\circ + \cos 40^\circ = 2 \cos\left(\frac{140^\circ + 40^\circ}{2}\right) \cos\left(\frac{140^\circ - 40^\circ}{2}\right)\)
\(= 2 \cos 90^\circ \cos 50^\circ\)
Since \(\cos 90^\circ = 0\):
\(= 2 \times 0 \times \cos 50^\circ = 0 = \text{RHS}\)
Hence, proved.
In simple words: Combine two cosine terms using the sum-to-product formula to generate a \(\cos 90^\circ\) term, which evaluates to zero.

Exam Tip: Look for angles that sum or difference into standard values like \(60^\circ\) or \(90^\circ\) when applying transformation formulas.

Question 10. Prove that \(\frac{\sin A + \sin 3A + \sin 5A + \sin 7A}{\cos A + \cos 3A + \cos 5A + \cos 7A} = \tan 4A\)
Answer: Grouping terms in the numerator and denominator by pairing the first with the fourth, and the second with the third:
Numerator: \((\sin 7A + \sin A) + (\sin 5A + \sin 3A)\)
Denominator: \((\cos 7A + \cos A) + (\cos 5A + \cos 3A)\)
Applying sum-to-product formulas:
For numerator:
\(\sin 7A + \sin A = 2 \sin 4A \cos 3A\)
\(\sin 5A + \sin 3A = 2 \sin 4A \cos A\)
So numerator becomes:
\(2 \sin 4A \cos 3A + 2 \sin 4A \cos A = 2 \sin 4A (\cos 3A + \cos A)\)
For denominator:
\(\cos 7A + \cos A = 2 \cos 4A \cos 3A\)
\(\cos 5A + \cos 3A = 2 \cos 4A \cos A\)
So denominator becomes:
\(2 \cos 4A \cos 3A + 2 \cos 4A \cos A = 2 \cos 4A (\cos 3A + \cos A)\)
Dividing numerator by denominator:
\(\text{LHS} = \frac{2 \sin 4A (\cos 3A + \cos A)}{2 \cos 4A (\cos 3A + \cos A)}\)
Canceling common terms \(2\) and \((\cos 3A + \cos A)\):
\(\text{LHS} = \frac{\sin 4A}{\cos 4A} = \tan 4A = \text{RHS}\)
Hence, proved.
In simple words: Group terms such that their sums give the same midpoint angle (4A), apply sum-to-product formulas, and factor out common terms.

Exam Tip: When pairing four terms, always pair terms whose sums of angles are equal (\(7A + A = 6A + 2A\) etc.).