CBSE Class 10 Mathematics Trigonometry Assignment Set 28

Type I: Very Short Answer Type Questions (1 Mark Each)

Question 1. Find the value of \( \dfrac{\tan 30^\circ}{\cot 60^\circ} \).
Answer:
\( \cot 60^\circ = \dfrac{1}{\sqrt{3}} = \tan 30^\circ \), so the value is 1.
In simple words: cot 60° and tan 30° are equal.

Teacher's Note:
1. tan(90° − θ) = cot θ.

Question 2. If \( \sin A = \dfrac{1}{2} \), then find the value of cot A.
Answer:
\( \sin A = \dfrac{1}{2} \Rightarrow A = 30^\circ \) and \( \cot 30^\circ = \) \( \sqrt{3} \).
In simple words: sin 30° = 1/2.

Teacher's Note:
1. Alternatively P = 1, H = 2, B = √3.

Question 3. Find the value of the expression \( [\operatorname{cosec}(75^\circ + \theta) - \sec(15^\circ - \theta) - \tan(55^\circ + \theta) + \cot(35^\circ - \theta)] \).
Answer:
\( \operatorname{cosec}(75^\circ + \theta) = \sec(15^\circ - \theta) \) and \( \cot(35^\circ - \theta) = \tan(55^\circ + \theta) \).
The expression is \( 0 \).
In simple words: Both pairs are complementary-angle identities.

Teacher's Note:
1. Check that the angles in each pair add to 90°.

Question 4. Given that \( \sin\theta = \dfrac{a}{b} \), then find \( \cos\theta \).
Answer:
\( \cos\theta = \sqrt{1 - \dfrac{a^2}{b^2}} = \) \( \dfrac{\sqrt{b^2 - a^2}}{b} \).
In simple words: Use sin² + cos² = 1.

Teacher's Note:
1. θ is acute, so cos θ is positive.

Question 5. If \( \triangle ABC \) is right angled at C, then find the value of \( \cos(A + B) \).
Answer:
\( A + B = 90^\circ \), so \( \cos(A + B) = \cos 90^\circ = \) 0.
In simple words: The two acute angles add up to 90°.

Teacher's Note:
1. Angle sum is 180° and C = 90°.

Question 6. Given that \( \sin\alpha = \dfrac{1}{2} \) and \( \cos\beta = \dfrac{1}{2} \), then find the value of \( (\alpha + \beta) \).
Answer:
\( \alpha = 30^\circ \) and \( \beta = 60^\circ \), so \( \alpha + \beta = \) 90°.
In simple words: Read the angles from the table.

Teacher's Note:
1. Both are acute angles.

Question 7. Find the value of the expression \( \left[\dfrac{\sin^2 22^\circ + \sin^2 68^\circ}{\cos^2 22^\circ + \cos^2 68^\circ} + \sin^2 63^\circ + \cos 63^\circ\sin 27^\circ\right] \).
Answer:
\( \sin 68^\circ = \cos 22^\circ \) and \( \cos 68^\circ = \sin 22^\circ \), so the fraction is \( \dfrac{\sin^2 22^\circ + \cos^2 22^\circ}{\cos^2 22^\circ + \sin^2 22^\circ} = 1 \).
\( \cos 63^\circ = \sin 27^\circ \), so \( \sin^2 63^\circ + \cos 63^\circ\sin 27^\circ = \sin^2 63^\circ + \sin^2 27^\circ = \sin^2 63^\circ + \cos^2 63^\circ = 1 \).
The value is \( 1 + 1 = \) 2.
In simple words: Convert 68° and 27° to their complements.

Teacher's Note:
1. Each pair collapses with sin² + cos² = 1.

Question 8. If \( 4\tan\theta = 3 \), then find \( \left(\dfrac{4\sin\theta - \cos\theta}{4\sin\theta + \cos\theta}\right) \).
Answer:
Divide the numerator and denominator by \( \cos\theta \): \( \dfrac{4\tan\theta - 1}{4\tan\theta + 1} = \dfrac{3 - 1}{3 + 1} = \) 1/2.
In simple words: Divide top and bottom by cos θ.

Teacher's Note:
1. Use 4 tan θ = 3 directly.

Question 9. If \( \sqrt{3}\tan\theta = 1 \), then find the value of \( \sin^2\theta - \cos^2\theta \).
Answer:
\( \tan\theta = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ \). \( \sin^2 30^\circ - \cos^2 30^\circ = \dfrac{1}{4} - \dfrac{3}{4} = \) −1/2.
In simple words: Find θ first.

Teacher's Note:
1. tan 30° = 1/√3.

Type II: Short Answer Type Questions (2 Marks Each)

Question 10. Prove that \( \sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta = 1 \).
Answer:
Let \( a = \sin^2\theta \) and \( b = \cos^2\theta \), so a + b = 1. \( a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 1 - 3ab \).
LHS \( = 1 - 3ab + 3ab = 1 \). Hence proved.
In simple words: Use a³ + b³ = (a + b)³ − 3ab(a + b).

Teacher's Note:
1. Here a + b = 1.

Question 11. If \( \sin\theta + \cos\theta = \sqrt{2} \), then prove that \( \tan\theta + \cot\theta = 2 \).
Answer:
Squaring: \( 1 + 2\sin\theta\cos\theta = 2 \Rightarrow \sin\theta\cos\theta = \dfrac{1}{2} \).
\( \tan\theta + \cot\theta = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{1/2} = 2 \). Hence proved.
In simple words: Square the given relation.

Teacher's Note:
1. The largest possible value of sin θ + cos θ is √2.

Question 12. Prove that \( \dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = 2\operatorname{cosec}\theta \).
Answer:
LHS \( = \dfrac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)} = \dfrac{2 + 2\cos\theta}{\sin\theta(1 + \cos\theta)} = \dfrac{2}{\sin\theta} = 2\operatorname{cosec}\theta \).
In simple words: Take the common denominator.

Teacher's Note:
1. sin² + cos² = 1 simplifies the numerator.

Question 13. Prove that \( (\sin\alpha + \cos\alpha)(\tan\alpha + \cot\alpha) = \sec\alpha + \operatorname{cosec}\alpha \).
Answer:
LHS \( = (\sin\alpha + \cos\alpha) \cdot \dfrac{1}{\sin\alpha\cos\alpha} = \dfrac{1}{\cos\alpha} + \dfrac{1}{\sin\alpha} = \sec\alpha + \operatorname{cosec}\alpha \).
In simple words: tan + cot = 1/(sin cos).

Teacher's Note:
1. Split the fraction at the end.

Question 14. Prove that \( 1 + \dfrac{\cot^2\alpha}{1 + \operatorname{cosec}\alpha} = \operatorname{cosec}\alpha \).
Answer:
\( \cot^2\alpha = \operatorname{cosec}^2\alpha - 1 = (\operatorname{cosec}\alpha - 1)(\operatorname{cosec}\alpha + 1) \).
LHS \( = 1 + (\operatorname{cosec}\alpha - 1) = \operatorname{cosec}\alpha \). Hence proved.
In simple words: Factorise cot² α.

Teacher's Note:
1. Cancel (1 + cosec α).

Question 15. Prove that \( \tan\theta + \tan(90^\circ - \theta) = \sec\theta\sec(90^\circ - \theta) \).
Answer:
LHS \( = \tan\theta + \cot\theta = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta} \). RHS \( = \sec\theta\operatorname{cosec}\theta = \dfrac{1}{\cos\theta\sin\theta} \). So LHS = RHS.
In simple words: Convert 90° − θ ratios first.

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

Question 16. If \( 2\sin^2\theta - \cos^2\theta = 2 \), then find the value of \( \theta \).
Answer:
\( 2\sin^2\theta - (1 - \sin^2\theta) = 2 \Rightarrow 3\sin^2\theta = 3 \Rightarrow \sin\theta = 1 \Rightarrow \theta = \) 90°.
In simple words: Replace cos² by 1 − sin².

Teacher's Note:
1. Check: 2 − 0 = 2.

Question 17. Show that \( \dfrac{\cos^2(45^\circ + \theta) + \cos^2(45^\circ - \theta)}{\tan(60^\circ + \theta)\tan(30^\circ - \theta)} = 1 \).
Answer:
\( \cos(45^\circ - \theta) = \sin(45^\circ + \theta) \), so the numerator is \( \cos^2(45^\circ + \theta) + \sin^2(45^\circ + \theta) = 1 \).
\( \tan(30^\circ - \theta) = \cot(60^\circ + \theta) \), so the denominator is \( \tan(60^\circ + \theta)\cot(60^\circ + \theta) = 1 \). The ratio is 1.
In simple words: Use complementary angles in both numerator and denominator.

Teacher's Note:
1. 45° + θ and 45° − θ are complementary.

Type III: Long Answer Type Questions (3 Marks Each)

Question 18. If \( \operatorname{cosec}\theta + \cot\theta = p \), then prove that \( \cos\theta = \dfrac{p^2 - 1}{p^2 + 1} \).
Answer:
\( \operatorname{cosec}\theta - \cot\theta = \dfrac{1}{p} \) (since \( \operatorname{cosec}^2\theta - \cot^2\theta = 1 \)).
Adding and subtracting: \( \operatorname{cosec}\theta = \dfrac{1}{2}\left(p + \dfrac{1}{p}\right) \) and \( \cot\theta = \dfrac{1}{2}\left(p - \dfrac{1}{p}\right) \).
\( \cos\theta = \dfrac{\cot\theta}{\operatorname{cosec}\theta} = \dfrac{p - \frac{1}{p}}{p + \frac{1}{p}} = \dfrac{p^2 - 1}{p^2 + 1} \). Hence proved.
In simple words: Find cosec θ − cot θ using the identity.

Teacher's Note:
1. Divide cot θ by cosec θ to get cos θ.

Question 19. Given that \( \sin\theta + 2\cos\theta = 1 \), then prove that \( 2\sin\theta - \cos\theta = 2 \).
Answer:
Squaring and adding: \( (\sin\theta + 2\cos\theta)^2 + (2\sin\theta - \cos\theta)^2 = 5(\sin^2\theta + \cos^2\theta) = 5 \).
So \( (2\sin\theta - \cos\theta)^2 = 5 - 1 = 4 \) and \( 2\sin\theta - \cos\theta = \pm 2 \).
For \( 0^\circ \le \theta \le 90^\circ \) we have \( \sin\theta + 2\cos\theta = 1 \) only when \( \theta = 90^\circ \) (the value is greater than 1 for smaller angles), and then \( 2\sin\theta - \cos\theta = 2 \). Hence proved.
In simple words: Square both expressions and add.

Teacher's Note:
1. Without the angle restriction the other sign is also possible.

Question 20. If \( a\sin\theta + b\cos\theta = c \), then prove that \( a\cos\theta - b\sin\theta = \pm\sqrt{a^2 + b^2 - c^2} \).
Answer:
\( (a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2 \).
So \( (a\cos\theta - b\sin\theta)^2 = a^2 + b^2 - c^2 \), that is \( a\cos\theta - b\sin\theta = \pm\sqrt{a^2 + b^2 - c^2} \). Hence proved.
In simple words: Square both and add; the cross terms cancel.

Teacher's Note:
1. Both signs are possible in general.

Question 21. If \( 1 + \sin^2\theta = 3\sin\theta\cos\theta \), then prove that \( \tan\theta = 1 \) or \( \dfrac{1}{2} \).
Answer:
Divide by \( \cos^2\theta \): \( \sec^2\theta + \tan^2\theta = 3\tan\theta \Rightarrow 1 + 2\tan^2\theta = 3\tan\theta \).
\( 2\tan^2\theta - 3\tan\theta + 1 = 0 \Rightarrow (2\tan\theta - 1)(\tan\theta - 1) = 0 \). So \( \tan\theta = 1 \) or \( \dfrac{1}{2} \). Hence proved.
In simple words: Divide every term by cos² θ.

Teacher's Note:
1. 1 = sin² + cos², so 1 + sin² divided by cos² is sec² + tan².

Type IV: Very Long Answer Type Questions (3 to 4 Marks Each)

Question 22. Prove that \( \dfrac{1 + \sec\theta - \tan\theta}{1 + \sec\theta + \tan\theta} = \dfrac{1 - \sin\theta}{\cos\theta} \).
Answer:
Write 1 as \( \sec^2\theta - \tan^2\theta \) in the numerator: \( \sec\theta - \tan\theta + (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = (\sec\theta - \tan\theta)(1 + \sec\theta + \tan\theta) \).
So LHS \( = \sec\theta - \tan\theta = \dfrac{1 - \sin\theta}{\cos\theta} = \) RHS. Hence proved.
In simple words: Replace the 1 on top by sec² − tan².

Teacher's Note:
1. The numerator factors with the denominator.

Question 23. If \( \tan\theta + \sec\theta = l \), then prove that \( \sec\theta = \dfrac{l^2 + 1}{2l} \).
Answer:
\( \sec\theta - \tan\theta = \dfrac{1}{l} \) (since \( \sec^2\theta - \tan^2\theta = 1 \)).
Adding: \( 2\sec\theta = l + \dfrac{1}{l} \Rightarrow \sec\theta = \dfrac{l^2 + 1}{2l} \). Hence proved.
In simple words: Find sec θ − tan θ first.

Teacher's Note:
1. Add the two equations.