CBSE Class 11 Mathematics Trigonometry Assignment Set 10

XI - Assignment No. 10 - Trigonometrical Functions

Note:- Questions based on \(\sin(A \pm B)\); \(\cos(A \pm B)\); \(\tan(A \pm B)\); \(\cot(A \pm B)\)

Question 1. Find the values of
(i) \(\sin 22^{\circ} \cos 38^{\circ} + \cos 22^{\circ} \sin 38^{\circ}\)
(ii) \(\sin 40^{\circ} \cos 10^{\circ} - \cos 40^{\circ} \sin 10^{\circ}\)
Answer:
(i) Using the formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\), where \(A = 22^{\circ}\) and \(B = 38^{\circ}\):
\(\sin 22^{\circ} \cos 38^{\circ} + \cos 22^{\circ} \sin 38^{\circ} = \sin(22^{\circ} + 38^{\circ})\)
\(= \sin(60^{\circ}) = \frac{\sqrt{3}}{2}\)

(ii) Using the formula \(\sin(A - B) = \sin A \cos B - \cos A \sin B\), where \(A = 40^{\circ}\) and \(B = 10^{\circ}\):
\(\sin 40^{\circ} \cos 10^{\circ} - \cos 40^{\circ} \sin 10^{\circ} = \sin(40^{\circ} - 10^{\circ})\)
\(= \sin(30^{\circ}) = \frac{1}{2}\)
In simple words: Apply compound angle formulas for sine sum and difference to simplify the expressions directly to standard trigonometric values.

Exam Tip: Memorize standard angle values and the exact expansion forms of compound angle formulas.

Question 2. Prove that: \(\cos(45^{\circ} - A) \cos(45^{\circ} - B) - \sin(45^{\circ} - A) \sin(45^{\circ} - B) = \sin(A + B)\)
Answer:
Let \(X = 45^{\circ} - A\) and \(Y = 45^{\circ} - B\). The given expression is of the form:
\(\cos X \cos Y - \sin X \sin Y = \cos(X + Y)\)
Substitute back the values of \(X\) and \(Y\):
\(\cos[(45^{\circ} - A) + (45^{\circ} - B)] = \cos(90^{\circ} - (A + B))\)
Using the co-function identity \(\cos(90^{\circ} - \theta) = \sin \theta\):
\(= \sin(A + B)\)
Hence proved.
In simple words: Use the cosine addition formula \(\cos(X+Y)\) and simplify the angles using complementary angle relations.

Exam Tip: Recognizing standard trigonometric identity expansions in variable substitution form saves time during exams.

Question 3. Prove that: \(\sin(n+1)A \sin(n-1)A + \cos(n+1)A \cos(n-1)A = \cos 2A\)
Answer:
Rearranging the terms using commutative property of addition:
\(= \cos(n+1)A \cos(n-1)A + \sin(n+1)A \sin(n-1)A\)
This is in the exact form of the cosine subtraction formula \(\cos(X - Y) = \cos X \cos Y + \sin X \sin Y\), where \(X = (n+1)A\) and \(Y = (n-1)A\):
\(= \cos[(n+1)A - (n-1)A]\)
Expand the angle inside the cosine function:
\(= \cos[nA + A - nA + A]\)
\(= \cos(2A)\)
Hence proved.
In simple words: Rearrange terms to match the cosine difference formula, then simplify the algebraic expression inside the angle.

Exam Tip: Always check if rearranging terms can directly reveal a standard compound angle identity.

Question 4. Prove that: \((\tan A + \tan B) \cos A \cos B = \sin(A + B)\)
Answer:
Start with the LHS and express \(\tan A\) and \(\tan B\) in terms of sine and cosine:
\(\text{LHS} = \left( \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} \right) \cos A \cos B\)
Take the LCM inside the parenthesis:
\(= \left( \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} \right) \cos A \cos B\)
Cancel out the common term \(\cos A \cos B\):
\(= \sin A \cos B + \cos A \sin B\)
Using the sine addition formula:
\(= \sin(A + B) = \text{RHS}\)
Hence proved.
In simple words: Convert tangent functions into sines and cosines, take a common denominator, and simplify.

Exam Tip: Converting tangents and cotangents to sines and cosines is a reliable universal technique for trigonometric proofs.

Question 5. If \(\sin A = \frac{8}{17}\) and \(\sin B = \frac{5}{13}\), find \(\sin(A - B)\)
Answer:
Given \(\sin A = \frac{8}{17}\), assuming acute angles unless specified otherwise, we find \(\cos A\):
\(\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}\)
Given \(\sin B = \frac{5}{13}\), we find \(\cos B\):
\(\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}\)
Now, use the formula for \(\sin(A - B)\):
\(\sin(A - B) = \sin A \cos B - \cos A \sin B\)
Substitute the values:
\(\sin(A - B) = \left(\frac{8}{17}\right) \left(\frac{12}{13}\right) - \left(\frac{15}{17}\right) \left(\frac{5}{13}\right)\)
\(= \frac{96}{221} - \frac{75}{221} = \frac{21}{221}\)
In simple words: Calculate missing cosine values using trigonometric identities, then apply the sine difference formula.

Exam Tip: Be careful with quadrants if angle ranges are specified; assume positive acute values if no quadrant is mentioned.

Question 6. Prove that: \(\frac{\sin(A + B)}{\sin(A - B)} = \frac{\tan A + \tan B}{\tan A - \tan B}\)
Answer:
Expand the numerator and denominator using compound angle formulas:
\(\text{LHS} = \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}\)
Divide every term in both the numerator and denominator by \(\cos A \cos B\):
\(= \frac{\frac{\sin A \cos B}{\cos A \cos B} + \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B}}\)
Simplify each fraction:
\(= \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}\)
Substitute \(\tan A = \frac{\sin A}{\cos A}\) and \(\tan B = \frac{\sin B}{\cos B}\):
\(= \frac{\tan A + \tan B}{\tan A - \tan B} = \text{RHS}\)
Hence proved.

In simple words: Expand numerator and denominator, then divide throughout by \(\cos A \cos B\) to convert into tangents.

Exam Tip: Dividing numerator and denominator by \(\cos A \cos B\) is a standard technique when converting sine/cosine combinations into tangent forms.

Question 7. Prove that: \(\cos\left(\frac{\pi}{4} + x\right) - \cos\left(\frac{\pi}{4} - x\right) = -\sqrt{2} \sin x\)
Answer:
Using the sum-to-product formula for cosines: \(\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)\)
Here, \(C = \frac{\pi}{4} + x\) and \(D = \frac{\pi}{4} - x\):
\(C + D = \left(\frac{\pi}{4} + x\right) + \left(\frac{\pi}{4} - x\right) = \frac{\pi}{2}\)
\(C - D = \left(\frac{\pi}{4} + x\right) - \left(\frac{\pi}{4} - x\right) = 2x\)
Substitute these into the formula:
\(\text{LHS} = -2 \sin\left(\frac{\pi/2}{2}\right) \sin\left(\frac{2x}{2}\right)\)
\(= -2 \sin\left(\frac{\pi}{4}\right) \sin x\)
Since \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
\(= -2 \left(\frac{1}{\sqrt{2}}\right) \sin x = -\sqrt{2} \sin x = \text{RHS}\)
Hence proved.
In simple words: Apply the standard factor formula \(\cos C - \cos D\) to simplify the difference of two cosines.

Exam Tip: Transformation formulas (sum-to-product and product-to-sum) are extremely helpful for simplifying trigonometric expressions involving sums and differences.

Question 8. Prove the following: \(\frac{\tan\left(\frac{\pi}{4} + x\right)}{\tan\left(\frac{\pi}{4} - x\right)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2\)
Answer:
Using the formula for \(\tan(A + B)\) and \(\tan(A - B)\) where \(A = \frac{\pi}{4}\) and \(B = x\):
Numerator: \(\tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan x}{1 - \tan\left(\frac{\pi}{4}\right) \tan x} = \frac{1 + \tan x}{1 - \tan x}\)
Denominator: \(\tan\left(\frac{\pi}{4} - x\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan x}{1 + \tan\left(\frac{\pi}{4}\right) \tan x} = \frac{1 - \tan x}{1 + \tan x}\)
Substitute both into the given fraction:
\(\text{LHS} = \frac{\frac{1 + \tan x}{1 - \tan x}}{\frac{1 - \tan x}{1 + \tan x}}\)
\(= \left(\frac{1 + \tan x}{1 - \tan x}\right) \times \left(\frac{1 + \tan x}{1 - \tan x}\right) = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2 = \text{RHS}\)
Hence proved.
In simple words: Expand numerator and denominator separately using the tangent sum and difference formulas, noting that \(\tan\left(\frac{\pi}{4}\right) = 1\).

Exam Tip: Remember the standard expansions for \(\tan\left(\frac{\pi}{4} + x\right)\) as they appear frequently in calculus and trigonometry problems.

Question 9. Show that: \(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\)
Answer:
Consider the identity involving angle addition: \(3x = 2x + x\)
Apply tangent on both sides:
\(\tan(3x) = \tan(2x + x)\)
Using the tangent addition formula:
\(\tan(3x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}\)
Cross-multiply to clear the denominator:
\(\tan(3x)(1 - \tan 2x \tan x) = \tan 2x + \tan x\)
Expand the left side:
\(\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x\)
Rearrange terms by moving \(\tan 3x \tan 2x \tan x\) to the right side and \((\tan 2x + \tan x)\) to the left side:
\(\tan 3x - \tan 2x - \tan x = \tan 3x \tan 2x \tan x\)
Hence proved.
In simple words: Start with the relation \(3x = 2x + x\), apply the tangent formula, and cross-multiply to rearrange terms.

Exam Tip: When proving product-to-difference relations for tangents, always start from the angle split identity like \(3x = 2x + x\) or \(A = B + C\).

Question 10. Show that: \(\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1\)
Answer:
Start with the angle relation \(3x = 2x + x\).
Apply cotangent on both sides:
\(\cot(3x) = \cot(2x + x)\)
Using the cotangent addition formula \(\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}\):
\(\cot(3x) = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}\)
Cross-multiply:
\(\cot(3x)(\cot x + \cot 2x) = \cot 2x \cot x - 1\)
Expand the left side:
\(\cot 3x \cot x + \cot 3x \cot 2x = \cot 2x \cot x - 1\)
Rearrange terms to isolate \(1\):
\(1 = \cot 2x \cot x - \cot 3x \cot 2x - \cot 3x \cot x\)
Rewriting in the standard order:
\(\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1\)
Hence proved.
In simple words: Begin with \(3x = 2x + x\), apply the cotangent addition formula, and cross-multiply to arrive at the identity.

Exam Tip: The cotangent addition formula is analogous to the tangent formula; be careful with the placement of terms in the numerator and denominator.