CBSE Class 10 Mathematics Trigonometry Assignment Set 26

Short Answer Type Questions

Question 1. Consider \( \triangle ABC \) in which \( \angle B = 90^\circ \), \( AB = 7 \) cm and \( AC - BC = 1 \) cm. Find the values of \( \sin C \) and \( \cos C \).

Answer:
Let \( AC = x \) cm. Then \( BC = (x - 1) \) cm. By the Pythagoras theorem, \( AB^2 + BC^2 = AC^2 \Rightarrow 49 + (x - 1)^2 = x^2 \Rightarrow 49 + x^2 - 2x + 1 = x^2 \Rightarrow 2x = 50 \Rightarrow x = 25 \).
So \( AC = 25 \) cm, \( BC = 24 \) cm and \( AB = 7 \) cm.
\( \sin C = \frac{AB}{AC} = \frac{7}{25} \) and \( \cos C = \frac{BC}{AC} = \frac{24}{25} \).
In simple words: Call the hypotenuse x, the other side is x − 1; the equation gives the 7, 24, 25 triangle.

Teacher's Note:
1. Draw the figure and mark all the given lengths.
2. Find the unknown sides with the Pythagoras theorem before writing any ratio.

Question 2. If \( \tan A = \frac{3}{4} \), find the value of \( \frac{1}{\sin A} + \frac{1}{\cos A} \).
Answer:
\( \tan A = \frac{3}{4} = \frac{BC}{AB} \). Let \( BC = 3k \) and \( AB = 4k \). Then \( AC = \sqrt{9k^2 + 16k^2} = 5k \).
\( \sin A = \frac{3k}{5k} = \frac{3}{5} \) and \( \cos A = \frac{4k}{5k} = \frac{4}{5} \).
\( \frac{1}{\sin A} + \frac{1}{\cos A} = \frac{5}{3} + \frac{5}{4} = \frac{20 + 15}{12} = \frac{35}{12} \).
In simple words: Find sin A and cos A from the 3, 4, 5 triangle and add their reciprocals.

Teacher's Note:
1. Draw the figure and mark all the given lengths.
2. Find the unknown sides with the Pythagoras theorem before writing any ratio.

Question 3. The rod AC of a TV disc antenna is fixed at right angles to the wall AB and a rod CD is supporting the disc as shown in the figure. If \( AC = 1.5 \) m long and \( CD = 3 \) m, find (i) \( \tan \theta \) (ii) \( \sec \theta + \operatorname{cosec} \theta \).

Answer:
In the right triangle CAD (right angle at A): \( AD^2 = CD^2 - AC^2 = 3^2 - (1.5)^2 = 9 - 2.25 = 6.75 \Rightarrow AD = \frac{3\sqrt{3}}{2} \) m.
(i) \( \tan \theta = \frac{AC}{AD} = \frac{1.5}{\frac{3\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \).
(ii) \( \sec \theta = \frac{CD}{AD} = \frac{3}{\frac{3\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \) and \( \operatorname{cosec} \theta = \frac{CD}{AC} = \frac{3}{1.5} = 2 \). So \( \sec \theta + \operatorname{cosec} \theta = \frac{2}{\sqrt{3}} + 2 = \frac{2 + 2\sqrt{3}}{\sqrt{3}} \).
In simple words: The rod CD is the hypotenuse; the angle θ is at D, so AC is opposite and AD is adjacent to it.

Teacher's Note:
1. Find AD first by the Pythagoras theorem.
2. Keep exact values such as 1/√3 instead of rounded decimals.

Question 4. If \( \sec A = \frac{26}{10} \), then find the value of \( \tan^2 A - \sin^2 A - \sin^4 A \sec^2 A \).
Answer:
\( \sec A = \frac{26}{10} = \frac{13}{5} = \frac{\text{Hypotenuse}}{\text{Base}} \). Let \( AC = 13k \) and \( AB = 5k \). Then \( BC = \sqrt{169k^2 - 25k^2} = 12k \).
\( \tan A = \frac{12}{5} \) and \( \sin A = \frac{12}{13} \).
\( \tan^2 A - \sin^2 A - \sin^4 A \sec^2 A = \frac{144}{25} - \frac{144}{169} - \frac{144^2}{169^2} \times \frac{169}{25} = \frac{144}{25} - \frac{144}{169} - \frac{20736}{4225} = \frac{24336 - 3600 - 20736}{4225} = 0 \).
In simple words: The result is 0 for every angle, because tan² A − sin² A = sin² A tan² A = sin⁴ A sec² A.

Teacher's Note:
1. Find sin A and tan A from the 5, 12, 13 triangle.
2. You can also check by the identity tan² A − sin² A = sin² A tan² A.

Question 5. If \( \sin(A + B) = \frac{\sqrt{3}}{2} \) and \( \sin(A - B) = \frac{1}{2} \); \( 0^\circ < A + B \le 90^\circ \); \( A > B \), find A and B.
Answer:
\( \sin(A + B) = \frac{\sqrt{3}}{2} = \sin 60^\circ \Rightarrow A + B = 60^\circ \) ....(i)
\( \sin(A - B) = \frac{1}{2} = \sin 30^\circ \Rightarrow A - B = 30^\circ \) ....(ii)
Adding (i) and (ii): \( 2A = 90^\circ \Rightarrow A = 45^\circ \). From (i): \( B = 15^\circ \).
In simple words: Convert each sine value to its angle and solve the two equations together.

Teacher's Note:
1. Write each given value as a standard angle.
2. Add and then subtract the two equations.

Question 6. If \( \sin \theta = \cos \theta \), then find the value of \( 2\tan \theta + \cos^2 \theta \).
Answer:
\( \sin \theta = \cos \theta \Rightarrow \tan \theta = 1 \Rightarrow \theta = 45^\circ \).
\( 2\tan \theta + \cos^2 \theta = 2 \times 1 + \left( \frac{1}{\sqrt{2}} \right)^2 = 2 + \frac{1}{2} = \frac{5}{2} \).
In simple words: The sine and cosine are equal only at 45°.

Teacher's Note:
1. Find the angle first and then substitute.
2. cos² 45° = 1/2.

Question 7. Evaluate: \( \frac{3\tan^2 30^\circ + \tan^2 60^\circ + \operatorname{cosec} 30^\circ - \tan 45^\circ}{\cot^2 45^\circ} \).
Answer:
\( = \frac{3\left( \frac{1}{\sqrt{3}} \right)^2 + (\sqrt{3})^2 + 2 - 1}{(1)^2} = \frac{3 \times \frac{1}{3} + 3 + 2 - 1}{1} = 1 + 3 + 2 - 1 = 5 \).
In simple words: Replace each ratio by its table value and add.

Teacher's Note:
1. Cosec 30° = 2 and tan 45° = cot 45° = 1.
2. Do not forget to square the tan 30° before multiplying by 3.

Question 8. Without using the trigonometric table, find the value of \( \frac{\sec(90^\circ - \theta)\operatorname{cosec} \theta - \tan(90^\circ - \theta)\cot \theta + (\cos^2 35^\circ + \cos^2 55^\circ)}{\tan 5^\circ \tan 15^\circ \tan 75^\circ \tan 85^\circ} \).
Answer:
\( \sec(90^\circ - \theta) = \operatorname{cosec} \theta \), \( \tan(90^\circ - \theta) = \cot \theta \) and \( \cos 55^\circ = \cos(90^\circ - 35^\circ) = \sin 35^\circ \).
Numerator \( = \operatorname{cosec}^2 \theta - \cot^2 \theta + (\cos^2 35^\circ + \sin^2 35^\circ) = 1 + 1 = 2 \).
Denominator \( = \tan 5^\circ \tan 15^\circ \cot 15^\circ \cot 5^\circ = 1 \) [using \( \tan 75^\circ = \cot 15^\circ \) and \( \tan 85^\circ = \cot 5^\circ \)].
The value is \( \frac{2}{1} = 2 \).
In simple words: Pairs of complementary angles turn the numerator into 1 + 1 and the denominator into 1.

Teacher's Note:
1. Use sec (90° − θ) = cosec θ and tan (90° − θ) = cot θ.
2. tan θ × cot θ = 1.

Question 9. Without using trigonometric tables, find the value of \( \frac{\sin^2 20^\circ + \sin^2 70^\circ}{\cos^2 20^\circ + \cos^2 70^\circ} + \frac{\sin \theta}{\tan \theta \sec \theta} + \frac{\cos \theta}{\cot \theta \operatorname{cosec} \theta} \).
Answer:
\( \sin 70^\circ = \cos 20^\circ \) and \( \cos 70^\circ = \sin 20^\circ \). The first term \( = \frac{\sin^2 20^\circ + \cos^2 20^\circ}{\cos^2 20^\circ + \sin^2 20^\circ} = \frac{1}{1} = 1 \).
\( \frac{\sin \theta}{\tan \theta \sec \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta}} = \cos^2 \theta \) and \( \frac{\cos \theta}{\cot \theta \operatorname{cosec} \theta} = \frac{\cos \theta}{\frac{\cos \theta}{\sin \theta} \times \frac{1}{\sin \theta}} = \sin^2 \theta \).
The value \( = 1 + \cos^2 \theta + \sin^2 \theta = 1 + 1 = 2 \).
In simple words: Each group of terms collapses to 1, giving 2.

Teacher's Note:
1. Use complementary angles for the first fraction.
2. Write the other two fractions in terms of sin θ and cos θ.

Question 10. A, B, C are interior angles of \( \triangle ABC \). Prove that \( \operatorname{cosec}\left( \frac{A + B}{2} \right) = \sec \frac{C}{2} \).
Answer:
In a triangle \( A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \Rightarrow \frac{A + B}{2} = 90^\circ - \frac{C}{2} \).
\( \operatorname{cosec}\left( \frac{A + B}{2} \right) = \operatorname{cosec}\left( 90^\circ - \frac{C}{2} \right) = \sec \frac{C}{2} \). Hence proved.
In simple words: Half of A + B is 90° minus half of C, and cosec (90° − x) = sec x.

Teacher's Note:
1. Use the angle sum property of a triangle.
2. Apply cosec (90° − θ) = sec θ.

Question 11. If \( \frac{\cos \alpha}{\cos \beta} = p \) and \( \frac{\cos \alpha}{\sin \beta} = q \), then prove that \( (p^2 + q^2)\cos^2 \beta = q^2 \).
Answer:
LHS \( = (p^2 + q^2)\cos^2 \beta = \left( \frac{\cos^2 \alpha}{\cos^2 \beta} + \frac{\cos^2 \alpha}{\sin^2 \beta} \right)\cos^2 \beta = \cos^2 \alpha\left( \frac{\sin^2 \beta + \cos^2 \beta}{\cos^2 \beta \sin^2 \beta} \right)\cos^2 \beta \).
\( = \frac{\cos^2 \alpha}{\sin^2 \beta} = \left( \frac{\cos \alpha}{\sin \beta} \right)^2 = q^2 = \) RHS. Hence proved.
In simple words: Substitute p and q, combine the fractions and use sin² β + cos² β = 1.

Teacher's Note:
1. Square p and q before adding.
2. The factor cos² β cancels with the denominator.

Question 12. If \( 4\tan \theta = 3 \), evaluate \( \frac{4\sin \theta - \cos \theta + 1}{4\sin \theta + \cos \theta - 1} \).
Answer:
\( \tan \theta = \frac{3}{4} \), so \( \sec^2 \theta = 1 + \frac{9}{16} = \frac{25}{16} \Rightarrow \sec \theta = \frac{5}{4} \).
Divide the numerator and the denominator by \( \cos \theta \): \( \frac{4\tan \theta - 1 + \sec \theta}{4\tan \theta + 1 - \sec \theta} = \frac{3 - 1 + \frac{5}{4}}{3 + 1 - \frac{5}{4}} = \frac{\frac{13}{4}}{\frac{11}{4}} = \frac{13}{11} \).
In simple words: Dividing by cos θ changes the expression into tan θ and sec θ, whose values are known.

Teacher's Note:
1. Find sec θ from sec² θ = 1 + tan² θ.
2. Divide every term by cos θ.