CBSE Class 11 Mathematics Trigonometry Assignment Set 12

Assignment No. 9 - Trigonometric Functions (Allied Angles)

Q1. Show that \(\tan(180^\circ + A) \cdot \sin(90^\circ + A) \cdot \sec(90^\circ - A) = 1\)
Answer: LHS = \(\tan(180^\circ + A) \cdot \sin(90^\circ + A) \cdot \sec(90^\circ - A)\)
Using allied angle formulas:
\(\tan(180^\circ + A) = \tan A\)
\(\sin(90^\circ + A) = \cos A\)
\(\sec(90^\circ - A) = \csc A\)
Substitute these values into the expression:
LHS = \(\tan A \cdot \cos A \cdot \csc A\)
Rewrite \(\tan A\) as \(\frac{\sin A}{\cos A}\) and \(\csc A\) as \(\frac{1}{\sin A}\):
LHS = \(\left(\frac{\sin A}{\cos A}\right) \cdot \cos A \cdot \left(\frac{1}{\sin A}\right)\)
LHS = \(1 =\) RHS
Hence proved.
In simple words: Convert each trigonometric function using standard allied angle rules and simplify into sine and cosine terms to get 1.

Exam Tip: Always memorize standard allied angle transformations for \(90^\circ \pm \theta\) and \(180^\circ \pm \theta\) before attempting proofs.

Q2. Prove that \(\frac{\cos \theta}{\sin(90^\circ + \theta)} + \frac{\sin(-\theta)}{\sin(180^\circ + \theta)} - \frac{\tan(90^\circ + \theta)}{\cot \theta} = 3\)
Answer: Simplify each term of the LHS using allied angle formulas:
1. \(\sin(90^\circ + \theta) = \cos \theta\)
2. \(\sin(-\theta) = -\sin \theta\)
3. \(\sin(180^\circ + \theta) = -\sin \theta\)
4. \(\tan(90^\circ + \theta) = -\cot \theta\)
Substitute these into the expression:
LHS = \(\frac{\cos \theta}{\cos \theta} + \frac{-\sin \theta}{-\sin \theta} - \frac{-\cot \theta}{\cot \theta}\)
LHS = \(1 + 1 - (-1)\)
LHS = \(1 + 1 + 1 = 3 =\) RHS
Hence proved.
In simple words: Simplify each individual fraction using basic trigonometric identities of allied angles to get \(1 + 1 + 1\).

Exam Tip: Be careful with negative signs when dealing with functions like \(\sin(-\theta)\) and \(\tan(90^\circ + \theta)\).

Q3. Find \(x\) from the trigonometrical equation \(\csc(90^\circ + A) + x \cos A \cot(90^\circ + A) = \sin(90^\circ + A)\) [Ans: \(\tan A\)]
Answer: Simplify the terms using allied angle formulas:
\(\csc(90^\circ + A) = \sec A\)
\(\cot(90^\circ + A) = -\tan A\)
\(\sin(90^\circ + A) = \cos A\)
Substitute these into the given equation:
\(\sec A + x \cos A (-\tan A) = \cos A\)
\(\frac{1}{\cos A} - x \cos A \left(\frac{\sin A}{\cos A}\right) = \cos A\)
\(\frac{1}{\cos A} - x \sin A = \cos A\)
Rearranging to solve for \(x \sin A\):
\(x \sin A = \frac{1}{\cos A} - \cos A\)
\(x \sin A = \frac{1 - \cos^2 A}{\cos A}\)
\(x \sin A = \frac{\sin^2 A}{\cos A}\)
Dividing both sides by \(\sin A\) (assuming \(\sin A \neq 0\)):
\(x = \frac{\sin A}{\cos A} = \tan A\)
In simple words: Substitute allied angle identities, simplify the algebraic expression, and isolate \(x\) to find the result.

Exam Tip: Convert secant, cosecant, tangent, and cotangent terms into sine and cosine to make algebraic manipulation easier.

Q4. Given \(\cot A = \tan(n-1)A\), find the value of \(A\) [Ans: \(\frac{\pi}{2n}\)]
Answer: We are given:
\(\cot A = \tan((n-1)A)\)
Convert \(\cot A\) to tangent using the relation \(\cot A = \tan\left(\frac{\pi}{2} - A \right)\):
\(\tan\left(\frac{\pi}{2} - A \right) = \tan((n-1)A)\)
Equating the general angles:
\(\frac{\pi}{2} - A = (n-1)A\)
\(\frac{\pi}{2} = (n-1)A + A\)
\(\frac{\pi}{2} = nA\)
Solving for \(A\):
\(A = \frac{\pi}{2n}\)
In simple words: Change cotangent to tangent using complementary angle relation \(90^\circ - A\), then equate the angles to solve for \(A\).

Exam Tip: Remember the co-function identity \(\cot \theta = \tan\left(\frac{\pi}{2} - \theta\right)\) to solve equations involving different trigonometric ratios.

Q5. Reduce to its simplest form \(\frac{\cos(90^\circ + \theta) \cdot \sec(-\theta) \cdot \tan(180^\circ - \theta)}{\sec(360^\circ - \theta) \cdot \sin(180^\circ + \theta) \cdot \cot(90^\circ - \theta)}\) [Ans: \(-1\)]
Answer: Simplify each term using allied angle formulas:
- Numerator:
\(\cos(90^\circ + \theta) = -\sin \theta\)
\(\sec(-\theta) = \sec \theta = \frac{1}{\cos \theta}\)
\(\tan(180^\circ - \theta) = -\tan \theta = -\frac{\sin \theta}{\cos \theta}\)
Product of numerator = \((-\sin \theta) \cdot \left(\frac{1}{\cos \theta}\right) \cdot \left(-\frac{\sin \theta}{\cos \theta}\right) = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta\)

- Denominator:
\(\sec(360^\circ - \theta) = \sec \theta = \frac{1}{\cos \theta}\)
\(\sin(180^\circ + \theta) = -\sin \theta\)
\(\cot(90^\circ - \theta) = \tan \theta = \frac{\sin \theta}{\cos \theta}\)
Product of denominator = \(\left(\frac{1}{\cos \theta}\right) \cdot (-\sin \theta) \cdot \left(\frac{\sin \theta}{\cos \theta}\right) = -\frac{\sin^2 \theta}{\cos^2 \theta} = -\tan^2 \theta\)

Now, divide numerator by denominator:
\(\frac{\tan^2 \theta}{-\tan^2 \theta} = -1\)
In simple words: Simplify numerator and denominator separately by applying standard allied angle formulas, then divide them to get \(-1\).

Exam Tip: Break complex expressions into individual trigonometric components and simplify step-by-step to avoid sign errors.

Q6. Prove that \(\cos 24^\circ + \cos 55^\circ + \cos 125^\circ + \cos 204^\circ + \cos 300^\circ = \frac{1}{2}\)
Answer: Consider the given expression:
LHS = \(\cos 24^\circ + \cos 55^\circ + \cos 125^\circ + \cos 204^\circ + \cos 300^\circ\)
Rewrite the angles using supplementary and allied angles:
\(\cos 204^\circ = \cos(180^\circ + 24^\circ) = -\cos 24^\circ\)
\(\cos 125^\circ = \cos(180^\circ - 55^\circ) = -\cos 55^\circ\)
\(\cos 300^\circ = \cos(360^\circ - 60^\circ) = \cos 60^\circ = \frac{1}{2}\)
Substitute these values back into LHS:
LHS = \(\cos 24^\circ + \cos 55^\circ + (-\cos 55^\circ) + (-\cos 24^\circ) + \frac{1}{2}\)
LHS = \((\cos 24^\circ - \cos 24^\circ) + (\cos 55^\circ - \cos 55^\circ) + \frac{1}{2}\)
LHS = \(0 + 0 + \frac{1}{2} = \frac{1}{2} =\) RHS
Hence proved.
In simple words: Express angles greater than \(90^\circ\) in terms of acute angles using \(180^\circ\) or \(360^\circ\) so that terms cancel out.

Exam Tip: Look for complementary or supplementary pairs that cancel each other out when simplified.

Q7. Find the value of \(\tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 88^\circ \tan 89^\circ\) [Ans: 1]
Answer: Given expression:
\(E = \tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 87^\circ \tan 88^\circ \tan 89^\circ\)
We know that \(\tan(90^\circ - \theta) = \cot \theta = \frac{1}{\tan \theta}\), so \(\tan \theta \cdot \tan(90^\circ - \theta) = 1\).
Pairing the terms from both ends:
\(\tan 1^\circ \cdot \tan 89^\circ = \tan 1^\circ \cdot \tan(90^\circ - 1^\circ) = \tan 1^\circ \cdot \cot 1^\circ = 1\)
\(\tan 2^\circ \cdot \tan 88^\circ = \tan 2^\circ \cdot \tan(90^\circ - 2^\circ) = \tan 2^\circ \cdot \cot 2^\circ = 1\)
Continuing this pairing for all intermediate terms up to \(\tan 45^\circ\) (where \(\tan 45^\circ = 1\)):
\(E = (1) \cdot (1) \cdot (1) \dots (1) = 1\)
In simple words: Pair each tangent with its complementary cotangent (since \(\tan(90^\circ - \theta) = \cot \theta\)), and since product of tan and cot is 1, the entire product evaluates to 1.

Exam Tip: For product series of tangents or sines, look for complementary angle relations where terms reciprocate and cancel out.

Q8. Show that \(\cos\left(\frac{3\pi}{2} + x\right) \cos(2\pi + x) \left[ \cot\left(\frac{3\pi}{2} - x\right) + \cot(2\pi + x) \right] = 1\)
Answer: Simplify each trigonometric term using allied angles (noting that \(\frac{3\pi}{2} = 270^\circ\) and \(2\pi = 360^\circ\)):
1. \(\cos\left(\frac{3\pi}{2} + x\right) = \sin x\)
2. \(\cos(2\pi + x) = \cos x\)
3. \(\cot\left(\frac{3\pi}{2} - x\right) = \tan x = \frac{\sin x}{\cos x}\)
4. \(\cot(2\pi + x) = \cot x = \frac{\cos x}{\sin x}\)
Substitute these into the LHS expression:
LHS = \(\sin x \cos x \left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right)\)
Take the LCM inside the bracket:
LHS = \(\sin x \cos x \left( \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \right)\)
Since \(\sin^2 x + \cos^2 x = 1\):
LHS = \(\sin x \cos x \left( \frac{1}{\sin x \cos x} \right)\)
LHS = \(1 =\) RHS
Hence proved.
In simple words: Convert each term using standard radian-based allied angle formulas, combine terms inside the bracket using a common denominator, and simplify using the identity \(\sin^2 x + \cos^2 x = 1\).

Exam Tip: Convert cotangent and tangent terms inside brackets into sine and cosine fractions to easily combine them using common denominators.