CBSE Class 11 Mathematics Trigonometry Assignment Set 09

XI - Assignment No. 12 - Trigonometrical Functions

Q1. Prove that \(\frac{\sin^2 A - \sin^2 B}{\cos^2 A - \sin^2 B} = \frac{\tan(A + B)}{\cot(A - B)}\)
Answer: LHS = \(\frac{\sin^2 A - \sin^2 B}{\cos^2 A - \sin^2 B}\)
Using the standard identities \(\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)\) and \(\cos^2 A - \sin^2 B = \cos(A + B)\cos(A - B)\), we get:
LHS = \(\frac{\sin(A + B)\sin(A - B)}{\cos(A + B)\cos(A - B)}\)
LHS = \(\left(\frac{\sin(A + B)}{\cos(A + B)}\right) \cdot \left(\frac{\sin(A - B)}{\cos(A - B)}\right)\)
LHS = \(\tan(A + B) \cdot \tan(A - B)\)
Since \(\tan(A - B) = \frac{1}{\cot(A - B)}\), we have:
LHS = \(\frac{\tan(A + B)}{\cot(A - B)} = \text{RHS}\)
In simple words: By applying compound angle formulas and converting the quotient of sines and cosines into tangents and cotangents, the left side equals the right side.

Exam Tip: Remember the useful property \(\cos^2 A - \sin^2 B = \cos(A + B)\cos(A - B)\) to quickly simplify expressions involving products and differences of squares.

Q2. Prove that \(\cos 2A \cos 2B + \sin^2(A - B) - \sin^2(A + B) = \cos(2A + 2B)\)
Answer: Consider the difference of squares term: \(\sin^2(A - B) - \sin^2(A + B)\)
Using the identity \(\sin^2 X - \sin^2 Y = \sin(X + Y)\sin(X - Y)\), let \(X = A - B\) and \(Y = A + B\):
\(X + Y = 2A\) and \(X - Y = -2B\)
So, \(\sin^2(A - B) - \sin^2(A + B) = \sin(2A)\sin(-2B) = -\sin 2A \sin 2B\)
Substitute this back into the LHS:
LHS = \(\cos 2A \cos 2B - \sin 2A \sin 2B\)
Using the cosine addition formula \(\cos(X + Y) = \cos X \cos Y - \sin X \sin Y\):
LHS = \(\cos(2A + 2B) = \text{RHS}\)
In simple words: Use the difference of two sines squared identity to simplify the middle terms, which then combines with the first term to form the cosine sum formula.

Exam Tip: Recognizing \(\sin^2 X - \sin^2 Y = \sin(X + Y)\sin(X - Y)\) saves considerable expansion steps.

Q3. Prove that \(\sin^2(n + 1)A - \sin^2 nA = \sin(2n + 1)A \sin A\)
Answer: Using the identity \(\sin^2 X - \sin^2 Y = \sin(X + Y)\sin(X - Y)\):
Here, \(X = (n + 1)A\) and \(Y = nA\)
Sum: \(X + Y = (n + 1)A + nA = (2n + 1)A\)
Difference: \(X - Y = (n + 1)A - nA = A\)
Therefore:
LHS = \(\sin((2n + 1)A) \sin A = \text{RHS}\)
In simple words: Apply the standard factorization formula for the difference of two squared sine terms directly.

Exam Tip: This is a standard trigonometric identity application; always check the sum and difference of the angles inside the squares.

Q4. Prove that \(\cos^2 C - \sin^2(C - D) = \cos D \cos(2C - D)\)
Answer: Using the identity \(\cos^2 X - \sin^2 Y = \cos(X + Y)\cos(X - Y)\):
Here, \(X = C\) and \(Y = C - D\)
Sum: \(X + Y = C + C - D = 2C - D\)
Difference: \(X - Y = C - (C - D) = D\)
Therefore:
LHS = \(\cos(2C - D) \cos D = \cos D \cos(2C - D) = \text{RHS}\)
In simple words: Apply the identity \(\cos^2 X - \sin^2 Y = \cos(X + Y)\cos(X - Y)\) by substituting \(X = C\) and \(Y = C - D\).

Exam Tip: Memorize the auxiliary formula \(\cos^2 X - \sin^2 Y = \cos(X + Y)\cos(X - Y)\) as it frequently simplifies mixed trigonometric squares.

Q5. Prove that \(\sin^2\left(\frac{\pi}{8} + \frac{A}{2}\right) - \sin^2\left(\frac{\pi}{8} - \frac{A}{2}\right) = \frac{1}{\sqrt{2}} \sin A\)
Answer: Using the identity \(\sin^2 X - \sin^2 Y = \sin(X + Y)\sin(X - Y)\):
Let \(X = \frac{\pi}{8} + \frac{A}{2}\) and \(Y = \frac{\pi}{8} - \frac{A}{2}\)
Sum: \(X + Y = \left(\frac{\pi}{8} + \frac{A}{2}\right) + \left(\frac{\pi}{8} - \frac{A}{2}\right) = \frac{2\pi}{8} = \frac{\pi}{4}\)
Difference: \(X - Y = \left(\frac{\pi}{8} + \frac{A}{2}\right) - \left(\frac{\pi}{8} - \frac{A}{2}\right) = A\)
Therefore:
LHS = \(\sin\left(\frac{\pi}{4}\right) \sin A\)
Since \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
LHS = \(\frac{1}{\sqrt{2}} \sin A = \text{RHS}\)
In simple words: Add and subtract the angles inside the sine squared terms to simplify using standard trigonometric values.

Exam Tip: Standard angles like \(\frac{\pi}{4}\) appear naturally when summing symmetric angle expressions.

Q6. If \(A + B = 45^\circ\), prove that \((1 + \tan A)(1 + \tan B) = 2\)
Answer: Given \(A + B = 45^\circ\), taking tangent on both sides:
\(\tan(A + B) = \tan 45^\circ = 1\)
Using the expansion formula for \(\tan(A + B)\):
\(\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1\)
\(\tan A + \tan B = 1 - \tan A \tan B\)
Rearranging terms:
\(\tan A + \tan B + \tan A \tan B = 1\)
Now expand the expression to prove: \((1 + \tan A)(1 + \tan B)\)
\(= 1 + \tan B + \tan A + \tan A \tan B\)
Substitute \(\tan A + \tan B + \tan A \tan B = 1\):
\(= 1 + 1 = 2 = \text{RHS}\)
In simple words: Expand the product, use the tangent sum formula for \(45^\circ\), and substitute the resulting relation to get 2.

Exam Tip: This is a classic result; whenever \(A + B = 45^\circ\) or \(\pi/4\), remember that \((1 + \tan A)(1 + \tan B) = 2\).

Q7. Prove that \(\tan 56^\circ = \frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ}\)
Answer: Consider the right-hand side:
RHS = \(\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ}\)
Divide both numerator and denominator by \(\cos 11^\circ\):
RHS = \(\frac{1 + \frac{\sin 11^\circ}{\cos 11^\circ}}{1 - \frac{\sin 11^\circ}{\cos 11^\circ}} = \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ}\)
We know that \(1 = \tan 45^\circ\), so substitute 1 in the numerator:
RHS = \(\frac{\tan 45^\circ + \tan 11^\circ}{1 - \tan 45^\circ \tan 11^\circ}\)
Using the tangent addition formula \(\tan(X + Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y}\):
RHS = \(\tan(45^\circ + 11^\circ) = \tan 56^\circ = \text{LHS}\)
In simple words: Divide numerator and denominator by cosine to convert the expression into tangent form, then use the angle sum formula.

Exam Tip: Dividing by \(\cos \theta\) to convert linear combinations of sine and cosine into tangent expressions is a powerful standard technique.

Q8. Find the value of \(\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ}\)
Answer: Using the tangent addition formula \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\), let \(A = 69^\circ\) and \(B = 66^\circ\):
Expression = \(\tan(69^\circ + 66^\circ)\)
Calculate the sum of angles:
\(69^\circ + 66^\circ = 135^\circ\)
So the expression becomes:
\(\tan 135^\circ = \tan(180^\circ - 45^\circ) = -\tan 45^\circ = -1\)
In simple words: Recognize the numerator and denominator pattern as the tangent sum formula, add the angles to get \(135^\circ\), and evaluate its trigonometric value.

Exam Tip: Always check if the compound angle simplifies to a standard angle like \(135^\circ\) or \(120^\circ\).

Q9. Prove that \(\tan 13A - \tan 9A - \tan 4A = \tan 13A \tan 9A \tan 4A\)
Answer: Start with the relation between the angles:
\(13A = 9A + 4A\)
Take tangent on both sides:
\(\tan 13A = \tan(9A + 4A)\)
Expand using the tangent addition formula:
\(\tan 13A = \frac{\tan 9A + \tan 4A}{1 - \tan 9A \tan 4A}\)
Cross-multiply to clear the denominator:
\(\tan 13A (1 - \tan 9A \tan 4A) = \tan 9A + \tan 4A\)
\(\tan 13A - \tan 13A \tan 9A \tan 4A = \tan 9A + \tan 4A\)
Rearranging the terms to isolate the product term:
\(\tan 13A - \tan 9A - \tan 4A = \tan 13A \tan 9A \tan 4A = \text{RHS}\)
In simple words: Express the largest angle as the sum of the two smaller angles, apply the tangent sum formula, and cross-multiply.

Exam Tip: When given three angles where one is the sum of the other two (like \(13A = 9A + 4A\)), always start by applying the tangent formula to that sum.

Q10. Prove that \(\tan 50^\circ = \tan 40^\circ + 2 \tan 10^\circ\)
Answer: Express \(50^\circ\) as \(40^\circ + 10^\circ\):
\(\tan 50^\circ = \tan(40^\circ + 10^\circ)\)
Expand using the tangent addition formula:
\(\tan 50^\circ = \frac{\tan 40^\circ + \tan 10^\circ}{1 - \tan 40^\circ \tan 10^\circ}\)
Cross-multiply:
\(\tan 50^\circ (1 - \tan 40^\circ \tan 10^\circ) = \tan 40^\circ + \tan 10^\circ\)
Since \(\tan 50^\circ = \tan(90^\circ - 40^\circ) = \cot 40^\circ = \frac{1}{\tan 40^\circ}\), substitute this in the equation:
\(\frac{1}{\tan 40^\circ} (1 - \tan 40^\circ \tan 10^\circ) = \tan 40^\circ + \tan 10^\circ\)
Multiply through by \(\tan 40^\circ\):
\(1 - \tan 40^\circ \tan 10^\circ = \tan^2 40^\circ + \tan 40^\circ \tan 10^\circ\)
Rearrange terms using \(1 - \tan^2 40^\circ = 2 \tan 40^\circ \cot 80^\circ\) or recognize complementary relations:
Alternatively, use \(\tan 50^\circ - \tan 40^\circ = \frac{\sin(50^\circ - 40^\circ)}{\cos 50^\circ \cos 40^\circ} = \frac{\sin 10^\circ}{\cos 50^\circ \cos 40^\circ}\)
Since \(\cos 50^\circ = \sin 40^\circ\), the denominator is \(\sin 40^\circ \cos 40^\circ = \frac{1}{2} \sin 80^\circ = \frac{1}{2} \cos 10^\circ\)
Thus, \(\tan 50^\circ - \tan 40^\circ = \frac{\sin 10^\circ}{\frac{1}{2} \cos 10^\circ} = 2 \tan 10^\circ\)
Therefore, \(\tan 50^\circ = \tan 40^\circ + 2 \tan 10^\circ = \text{RHS}\)
In simple words: Express \(\tan 50^\circ - \tan 40^\circ\) in terms of sines and cosines, simplify the denominator using double angle formulas, and rearrange.

Exam Tip: Converting tangents into sine and cosine is extremely helpful when proving difference identities involving complementary angles.