CBSE Class 11 Mathematics Trigonometry Assignment Set 11

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:
For part (i):
Using the formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\), let \(A = 22^\circ\) and \(B = 38^\circ\).
\(\sin 22^\circ \cos 38^\circ + \cos 22^\circ \sin 38^\circ = \sin(22^\circ + 38^\circ)\)
\(= \sin(60^\circ)\)
\(= \frac{\sqrt{3}}{2}\)
For part (ii):
Using the formula \(\sin(A - B) = \sin A \cos B - \cos A \sin B\), let \(A = 40^\circ\) 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: We applied the standard sine addition and subtraction angle formulas to simplify the expressions to known trigonometric values.

Exam Tip: Memorize the compound angle formulas for sine, cosine, and tangent thoroughly as they frequently appear in trigonometric simplification problems.

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 \(\cos X \cos Y - \sin X \sin Y\), which is equal to \(\cos(X + Y)\).
Substituting \(X\) and \(Y\):
\(\cos(X + Y) = \cos((45^\circ - A) + (45^\circ - B))\)
\(= \cos(90^\circ - (A + B))\)
Using the co-function identity \(\cos(90^\circ - \theta) = \sin \theta\):
\(= \sin(A + B)\)
Hence proved.
In simple words: We used the cosine addition formula in reverse along with a complementary angle identity to get the required result.

Exam Tip: Recognizing standard trigonometric expansions like \(\cos X \cos Y - \sin X \sin Y = \cos(X + Y)\) saves considerable time.

Question 3. Prove that: \(\sin(n + 1)A \sin(n - 1)A + \cos(n + 1)A \cos(n - 1)A = \cos 2A\)
Answer:
Let \(X = (n + 1)A\) and \(Y = (n - 1)A\).
The given expression is \(\cos X \cos Y + \sin X \sin Y\), which is equal to \(\cos(X - Y)\).
Substituting \(X\) and \(Y\):
\(\cos(X - Y) = \cos((n + 1)A - (n - 1)A)\)
\(= \cos(nA + A - nA + A)\)
\(= \cos(2A)\)
Hence proved.
In simple words: Rearranging the terms to fit the cosine difference formula \(\cos(X - Y) = \cos X \cos Y + \sin X \sin Y\) directly yields the answer after algebraic simplification of the angles.

Exam Tip: Always check if the expression matches the standard cosine subtraction expansion when cosines and sines are multiplied together with mixed angle terms.

Question 4. Prove that: \((\tan A + \tan B) \cos A \cos B = \sin(A + B)\)
Answer:
Starting with the Left Hand Side (LHS):
\(\text{LHS} = (\tan A + \tan B) \cos A \cos B\)
Expressing tangent in terms of sine and cosine: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
\(\text{LHS} = \left(\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}\right) \cos A \cos B\)
Taking the common denominator inside the parenthesis:
\(\text{LHS} = \left(\frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B}\right) \cos A \cos B\)
Canceling out \(\cos A \cos B\) from numerator and denominator:
\(\text{LHS} = \sin A \cos B + \cos A \sin B\)
Using the sine addition formula, this equals \(\sin(A + B)\), which is the Right Hand Side (RHS).
Hence proved.
In simple words: Converting tangents into sines and cosines makes it easy to simplify the expression by cross-multiplication and cancellation.

Exam Tip: When stuck with trigonometric identities involving tangents and cotangents, converting everything to sines and cosines is a reliable universal strategy.

Question 5. If \(\sin A = \frac{8}{17}\) and \(\sin B = \frac{5}{13}\), find \(\sin(A - B)\)
Answer:
Given \(\sin A = \frac{8}{17}\). Since \(\cos A = \sqrt{1 - \sin^2 A}\):
\(\cos A = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}\)
Given \(\sin B = \frac{5}{13}\). Since \(\cos B = \sqrt{1 - \sin^2 B}\):
\(\cos B = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}\)
The formula for \(\sin(A - B)\) is:
\(\sin(A - B) = \sin A \cos B - \cos A \sin B\)
Substituting the values:
\(\sin(A - B) = \left(\frac{8}{17}\right)\left(\frac{12}{13}\right) - \left(\frac{15}{17}\right)\left(\frac{5}{13}\right)\)
\(\sin(A - B) = \frac{96}{221} - \frac{75}{221} = \frac{21}{221}\)
In simple words: We first found the missing cosine values using the Pythagorean identity and then substituted all values into the sine difference formula.

Exam Tip: Always determine the signs of the square roots based on the given quadrants for angles A and B (assuming acute angles unless specified otherwise).

Question 6. Prove that: \(\frac{\sin(A + B)}{\sin(A - B)} = \frac{\tan A + \tan B}{\tan A - \tan B}\)
Answer:
Starting with the Left Hand Side (LHS):
\(\text{LHS} = \frac{\sin(A + B)}{\sin(A - B)}\)
Expanding using compound angle formulas:
\(\text{LHS} = \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}\)
Dividing every term in the numerator and the denominator by \(\cos A \cos B\):
\(\text{LHS} = \frac{\frac{\sin A \cos B}{\cos A \cos B} + \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B}}\)
Simplifying the fractions using \(\frac{\sin \theta}{\cos \theta} = \tan \theta\):
\(\text{LHS} = \frac{\tan A + \tan B}{\tan A - \tan B} = \text{RHS}\)
Hence proved.
In simple words: Expanding both sine terms and dividing each term by \(\cos A \cos B\) directly converts sines and cosines into tangents.

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

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 \(D = \frac{\pi}{4} - x\).
Calculate \(\frac{C + D}{2}\):
\(\frac{C + D}{2} = \frac{\left(\frac{\pi}{4} + x\right) + \left(\frac{\pi}{4} - x\right)}{2} = \frac{\frac{\pi}{2}}{2} = \frac{\pi}{4}\)
Calculate \(\frac{C - D}{2}\):
\(\frac{C - D}{2} = \frac{\left(\frac{\pi}{4} + x\right) - \left(\frac{\pi}{4} - x\right)}{2} = \frac{2x}{2} = x\)
Substituting back into the formula:
\(\cos\left(\frac{\pi}{4} + x\right) - \cos\left(\frac{\pi}{4} - x\right) = -2 \sin\left(\frac{\pi}{4}\right) \sin(x)\)
Since \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
\(= -2 \left(\frac{1}{\sqrt{2}}\right) \sin x = -\sqrt{2} \sin x\)
Hence proved.
In simple words: Applying the standard sum-to-product trigonometric transformation simplifies the difference of two cosines into a single product containing sine terms.

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 \(B = x\):
\(\tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan x}{1 - \tan\left(\frac{\pi}{4}\right) \tan x} = \frac{1 + \tan x}{1 - \tan x}\)
Similarly:
\(\tan\left(\frac{\pi}{4} - x\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan x}{1 + \tan\left(\frac{\pi}{4}\right) \tan x} = \frac{1 - \tan x}{1 + \tan x}\)
Now substituting these 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: Expanding both the numerator and denominator using the tangent addition and subtraction formulas with \(\tan\left(\frac{\pi}{4}\right) = 1\) leads directly to the squared ratio.

Exam Tip: Remember that \(\tan\left(\frac{\pi}{4} \pm x\right) = \frac{1 \pm \tan x}{1 \mp \tan x}\) is a very useful standard result to memorize directly.

Question 9. Show that: \(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\)
Answer:
Consider the identity involving the angle addition: \(3x = 2x + x\).
Taking 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-multiplying to clear the denominator:
\(\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x\)
\(\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x\)
Rearranging the terms to isolate the product term:
\(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\)
Hence proved.
In simple words: By starting from the basic relation \(3x = 2x + x\) and applying the tangent addition formula, we rearrange the algebraic terms to prove the identity.

Exam Tip: Splitting a multiple angle (like \(3x\) into \(2x + x\)) is a standard technique for proving product-to-difference identities involving tangents.

Question 10. Show that: \(\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1\)
Answer:
Consider the relation between angles: \(3x = 2x + x\).
Taking cotangent on both sides or using the cotangent identity for \(\cot(A + B)\):
\(\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}\)
Applying this for \(\cot 3x = \cot(2x + x)\):
\(\cot 3x = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}\)
Cross-multiplying:
\(\cot 3x (\cot x + \cot 2x) = \cot 2x \cot x - 1\)
\(\cot 3x \cot x + \cot 3x \cot 2x = \cot 2x \cot x - 1\)
Rearranging the terms to one side:
\(1 = \cot 2x \cot x - \cot 3x \cot 2x - \cot 3x \cot x\)
Rewriting the left and right sides:
\(\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1\)
Hence proved.
In simple words: Similar to Question 9, we start with \(\cot(2x + x)\) and use cross-multiplication to arrive at the cotangent product identity.

Exam Tip: Practice both tangent and cotangent expansion forms as they are mirror images and frequently tested in trigonometric proofs.