CBSE Class 10 Mathematics Trigonometry Assignment Set 27

Long Answer Type Questions

Question 1. In the given figure, \( AD = DB \) and \( \angle B \) is a right angle, with \( AC = b \) and \( AB = a \). Determine (i) \( \sin \theta \) (ii) \( \cos \theta \) (iii) \( \sin^2 \theta + \cos^2 \theta \).

Answer:
Since \( AD = DB \) and \( AD + DB = AB = a \), we get \( AD = DB = \frac{a}{2} \).
In right \( \triangle ABC \): \( AC^2 = AB^2 + BC^2 \Rightarrow b^2 = a^2 + BC^2 \Rightarrow BC = \sqrt{b^2 - a^2} \).
In right \( \triangle BCD \): \( CD^2 = BC^2 + BD^2 = b^2 - a^2 + \frac{a^2}{4} = \frac{4b^2 - 3a^2}{4} \Rightarrow CD = \frac{\sqrt{4b^2 - 3a^2}}{2} \).
(i) \( \sin \theta = \frac{BD}{CD} = \frac{a/2}{\frac{\sqrt{4b^2 - 3a^2}}{2}} = \frac{a}{\sqrt{4b^2 - 3a^2}} \).
(ii) \( \cos \theta = \frac{BC}{CD} = \frac{\sqrt{b^2 - a^2}}{\frac{\sqrt{4b^2 - 3a^2}}{2}} = \frac{2\sqrt{b^2 - a^2}}{\sqrt{4b^2 - 3a^2}} \).
(iii) \( \sin^2 \theta + \cos^2 \theta = \frac{a^2}{4b^2 - 3a^2} + \frac{4(b^2 - a^2)}{4b^2 - 3a^2} = \frac{a^2 + 4b^2 - 4a^2}{4b^2 - 3a^2} = \frac{4b^2 - 3a^2}{4b^2 - 3a^2} = 1 \).
In simple words: Find BC and CD with the Pythagoras theorem; the identity in part (iii) then checks the work.

Teacher's Note:
1. Draw the figure and write the given lengths on it.
2. Use the Pythagoras theorem to get each missing side.

Question 2. If \( \sqrt{3}\cot^2 \theta - 4\cot \theta + \sqrt{3} = 0 \), then find the value of \( \cot^2 \theta + \tan^2 \theta \).
Answer:
Let \( \cot \theta = x \). Then \( \sqrt{3}x^2 - 4x + \sqrt{3} = 0 \Rightarrow \sqrt{3}x^2 - 3x - x + \sqrt{3} = 0 \Rightarrow \sqrt{3}x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0 \Rightarrow (x - \sqrt{3})(\sqrt{3}x - 1) = 0 \).
So \( x = \sqrt{3} \) or \( x = \frac{1}{\sqrt{3}} \), i.e. \( \cot \theta = \sqrt{3} \) or \( \frac{1}{\sqrt{3}} \), giving \( \theta = 30^\circ \) or \( 60^\circ \).
If \( \theta = 30^\circ \): \( \cot^2 30^\circ + \tan^2 30^\circ = (\sqrt{3})^2 + \left( \frac{1}{\sqrt{3}} \right)^2 = 3 + \frac{1}{3} = \frac{10}{3} \). If \( \theta = 60^\circ \) the two terms are interchanged and the value is again \( \frac{10}{3} \).
Hence \( \cot^2 \theta + \tan^2 \theta = \frac{10}{3} \).
In simple words: Treat cot θ as x, factorise the quadratic and evaluate; both roots give the same answer.

Teacher's Note:
1. Split the middle term 4x as 3x + x.
2. Both angles 30° and 60° give the same value, so the answer is unique.

Question 3. In a \( \triangle ABC \) right angled at C, if \( \tan A = \frac{1}{\sqrt{3}} \) and \( \tan B = \sqrt{3} \). Show that \( \sin A \cos B + \cos A \sin B = 1 \).
Answer:
\( \tan A = \frac{1}{\sqrt{3}} = \tan 30^\circ \Rightarrow A = 30^\circ \) and \( \tan B = \sqrt{3} = \tan 60^\circ \Rightarrow B = 60^\circ \).
\( \sin A \cos B + \cos A \sin B = \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ = \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 \). Hence shown.
In simple words: Find the angles from the tangent values and substitute the table values.

Teacher's Note:
1. Compare tan values with the standard angles.
2. Remember that A + B = 90° here, so the expression is sin (A + B) = sin 90° = 1.

Question 4. Evaluate: \( \tan^2 30^\circ \sin 30^\circ + \cos 60^\circ \sin^2 90^\circ \tan^2 60^\circ - 2\tan 45^\circ \cos^2 0^\circ \sin 90^\circ \).
Answer:
\( = \left( \frac{1}{\sqrt{3}} \right)^2 \times \frac{1}{2} + \frac{1}{2} \times (1)^2 \times (\sqrt{3})^2 - 2 \times 1 \times 1^2 \times 1 \)
\( = \frac{1}{3} \times \frac{1}{2} + \frac{1}{2} \times 3 - 2 = \frac{1}{6} + \frac{3}{2} - 2 = \frac{1 + 9 - 12}{6} = -\frac{2}{6} = -\frac{1}{3} \).
In simple words: Write each table value, multiply term by term and combine the fractions.

Teacher's Note:
1. Use cos 0° = 1 and sin 90° = 1.
2. Combine the three terms over the common denominator 6.

Question 5. Prove that \( \frac{\sec^2 54^\circ - \cot^2 36^\circ}{\operatorname{cosec}^2 57^\circ - \tan^2 33^\circ} + 2\sin^2 38^\circ \sec^2 52^\circ - 2\sin^2 45^\circ = 2 \).
Answer:
\( \sec 54^\circ = \sec(90^\circ - 36^\circ) = \operatorname{cosec} 36^\circ \), \( \operatorname{cosec} 57^\circ = \operatorname{cosec}(90^\circ - 33^\circ) = \sec 33^\circ \) and \( \sec 52^\circ = \sec(90^\circ - 38^\circ) = \operatorname{cosec} 38^\circ \).
LHS \( = \frac{\operatorname{cosec}^2 36^\circ - \cot^2 36^\circ}{\sec^2 33^\circ - \tan^2 33^\circ} + 2\sin^2 38^\circ \operatorname{cosec}^2 38^\circ - 2\left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{1} + 2 - 1 = 2 = \) RHS. Hence proved.
In simple words: Complementary angles make the two fractions simple: the first is 1 and the second term is 2.

Teacher's Note:
1. Use sec (90° − θ) = cosec θ.
2. Use cosec² θ − cot² θ = 1 and sec² θ − tan² θ = 1.

Question 6. If \( \operatorname{cosec} \theta - \sin \theta = a \) and \( \sec \theta - \cos \theta = b \), prove that \( a^2 b^2 (a^2 + b^2 + 3) = 1 \).
Answer:
\( a = \operatorname{cosec} \theta - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \) and \( b = \sec \theta - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \).
\( a^2 b^2 = \frac{\cos^4 \theta}{\sin^2 \theta} \times \frac{\sin^4 \theta}{\cos^2 \theta} = \sin^2 \theta \cos^2 \theta \).
\( a^2 + b^2 + 3 = \frac{\cos^4 \theta}{\sin^2 \theta} + \frac{\sin^4 \theta}{\cos^2 \theta} + 3 = \frac{\cos^6 \theta + \sin^6 \theta + 3\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \).
Now \( \cos^6 \theta + \sin^6 \theta + 3\sin^2 \theta \cos^2 \theta = (\cos^2 \theta)^3 + (\sin^2 \theta)^3 + 3\sin^2 \theta \cos^2 \theta(\sin^2 \theta + \cos^2 \theta) = (\cos^2 \theta + \sin^2 \theta)^3 = 1 \).
Therefore \( a^2 b^2 (a^2 + b^2 + 3) = \sin^2 \theta \cos^2 \theta \times \frac{1}{\sin^2 \theta \cos^2 \theta} = 1 \). Hence proved.
In simple words: Write a and b in sine and cosine; the product turns into the cube of (sin² θ + cos² θ), which is 1.

Teacher's Note:
1. Use the identity a³ + b³ + 3ab(a + b) = (a + b)³.
2. Here a = cos² θ, b = sin² θ and a + b = 1.

Question 7. Prove that \( \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta} \).
Answer:
LHS: divide the numerator and the denominator by \( \cos \theta \): \( \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta} \).
Replace 1 in the numerator by \( \sec^2 \theta - \tan^2 \theta \): numerator \( = (\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = (\tan \theta + \sec \theta)[1 - \sec \theta + \tan \theta] \).
So LHS \( = \frac{(\tan \theta + \sec \theta)(\tan \theta + 1 - \sec \theta)}{\tan \theta + 1 - \sec \theta} = \tan \theta + \sec \theta \).
RHS \( = \frac{1}{\sec \theta - \tan \theta} \times \frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta} = \frac{\sec \theta + \tan \theta}{\sec^2 \theta - \tan^2 \theta} = \sec \theta + \tan \theta \).
LHS = RHS. Hence proved.
In simple words: Divide by cos θ, use sec² θ − tan² θ = 1 to factorise, and rationalise the right side.

Teacher's Note:
1. Rationalise 1/(sec θ − tan θ) by multiplying with (sec θ + tan θ).
2. Both sides finally become sec θ + tan θ.