2021-2022 Trigonometry Questions and Solutions
1.Given that cos θ = √3/2, then the value of \(\frac{\csc^2 \theta – \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta}\) is [Term I CBSE 2021-2022]
(a) -1
(b) 1
(c) \(\frac{1}{2}\)
(d) \(-\frac{1}{2}\)
View Answer
Solution:
Given: cos θ = √3/2 ⇒ θ = 30°
sin θ = 1/2, csc θ = 2, sec θ = 2/√3
csc²θ = 4, sec²θ = 4/3
Numerator: csc²θ – sec²θ = 4 – 4/3 = 8/3
Denominator: csc²θ + sec²θ = 4 + 4/3 = 16/3
Value = (8/3)/(16/3) = 1/2
Answer: (c) \(\frac{1}{2}\)
2. \(\frac{1}{\csc \theta (1 – \cot \theta)} + \frac{1}{\sec \theta (1 – \tan \theta)}\) is equal to [Term I CBSE 2021-2022]
(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ – cos θ
View Answer
Solution:
Simplify each term:
First term: \(\frac{1}{\csc \theta (1 – \cot \theta)} = \frac{\sin \theta}{1 – \cot \theta}\)
Second term: \(\frac{1}{\sec \theta (1 – \tan \theta)} = \frac{\cos \theta}{1 – \tan \theta}\)
Convert to sin and cos:
= \(\frac{\sin \theta}{1 – \frac{\cos \theta}{\sin \theta}} + \frac{\cos \theta}{1 – \frac{\sin \theta}{\cos \theta}}\)
= \(\frac{\sin \theta}{\frac{\sin \theta – \cos \theta}{\sin \theta}} + \frac{\cos \theta}{\frac{\cos \theta – \sin \theta}{\cos \theta}}\)
= \(\frac{\sin^2 \theta}{\sin \theta – \cos \theta} + \frac{\cos^2 \theta}{\cos \theta – \sin \theta}\)
= \(\frac{\sin^2 \theta – \cos^2 \theta}{\sin \theta – \cos \theta}\)
= \(\frac{(\sin \theta – \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta – \cos \theta}\)
= sin θ + cos θ
Answer: (c) sin θ + cos θ
3. If tan θ + cot θ = \(\frac{4\sqrt{3}}{3}\), then find the value of tan²θ + cot²θ [Term I CBSE 2021-2022]
View Answer
Solution:
We know: (tan θ + cot θ)² = tan²θ + cot²θ + 2tan θ cot θ
Since tan θ cot θ = 1, we have:
\(\left(\frac{4\sqrt{3}}{3}\right)^2 = \tan^2 \theta + \cot^2 \theta + 2\)
\(\frac{16 \times 3}{9} = \tan^2 \theta + \cot^2 \theta + 2\)
\(\frac{48}{9} = \tan^2 \theta + \cot^2 \theta + 2\)
\(\frac{16}{3} = \tan^2 \theta + \cot^2 \theta + 2\)
\(\tan^2 \theta + \cot^2 \theta = \frac{16}{3} – 2 = \frac{16}{3} – \frac{6}{3} = \frac{10}{3}\)
Answer: \(\frac{10}{3}\)
4.Given that sinα = √3/2 and tanβ = 1/√3, then the value of cos(α – β) is [Term I CBSE 2021-2022]
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(\frac{1}{2}\)
(c) 0
(d) \(\frac{1}{\sqrt{2}}\)
View Answer
Solution:
sinα = √3/2 ⇒ α = 60°
tanβ = 1/√3 ⇒ β = 30°
cos(α – β) = cos(60° – 30°) = cos 30° = √3/2
Answer: (a) \(\frac{\sqrt{3}}{2}\)
5.The value of θ for which 2 sin 2θ = 1, is [Term I CBSE 2021-2022]
(a) 15°
(b) 30°
(c) 45°
(d) 60°
View Answer
Solution:
2 sin 2θ = 1 ⇒ sin 2θ = 1/2
We know sin 30° = 1/2, so 2θ = 30° ⇒ θ = 15°
Answer: (a) 15°
6.Given that sin θ = p/q, tan θ is equal to [Term I CBSE 2021-2022]
(a) \(\frac{p}{\sqrt{p^2 – q^2}}\)
(b) \(\frac{q}{\sqrt{p^2 – q^2}}\)
(c) \(\frac{p}{\sqrt{q^2 – p^2}}\)
(d) \(\frac{q}{\sqrt{q^2 – p^2}}\)
View Answer
Solution:
sin θ = p/q
cos θ = √(1 – sin²θ) = √(1 – p²/q²) = √((q² – p²)/q²) = √(q² – p²)/q
tan θ = sin θ/cos θ = (p/q)/(√(q² – p²)/q) = p/√(q² – p²)
Answer: (c) \(\frac{p}{\sqrt{q^2 – p^2}}\)
7.The simplest form of √[(1 – cos²θ)(1 + tan²θ)] is [Term I CBSE 2021-2022]
(a) cos θ
(b) sin θ
(c) cot θ
(d) tan θ
View Answer
Solution:
√[(1 – cos²θ)(1 + tan²θ)] = √[sin²θ × sec²θ]
= √[sin²θ × 1/cos²θ] = √[tan²θ] = |tan θ|
For 0° ≤ θ ≤ 90°, tan θ ≥ 0, so = tan θ
Answer: (d) tan θ
8.If sin²θ + sinθ = 1, then the value of cos²θ + cos⁴θ is [Term I CBSE 2021-2022]
(a) -1
(b) 1
(c) 0
(d) 2
View Answer
Solution:
Given: sin²θ + sinθ = 1 ⇒ sin²θ = 1 – sinθ
We know sin²θ + cos²θ = 1 ⇒ cos²θ = 1 – sin²θ = sinθ
Now cos²θ + cos⁴θ = cos²θ + (cos²θ)² = sinθ + sin²θ = 1
Answer: (b) 1
9.If 3 sin A = 1, then find the value of sec A [Term I CBSE 2021-2022]
View Answer
Solution:
3 sin A = 1 ⇒ sin A = 1/3
cos A = √(1 – sin²A) = √(1 – 1/9) = √(8/9) = 2√2/3
sec A = 1/cos A = 3/(2√2) = (3√2)/4
Answer: \(\frac{3\sqrt{2}}{4}\)
10.Show that: \(\frac{1 + \cot^2 \theta}{1 + \tan^2 \theta} = \cot^2 \theta\) [Term I CBSE 2021-2022]
View Answer
Solution:
LHS = \(\frac{1 + \cot^2 \theta}{1 + \tan^2 \theta} = \frac{\csc^2 \theta}{\sec^2 \theta} = \frac{1/\sin^2 \theta}{1/\cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta\) = RHS
Hence proved.
11. If 4 tan β = 3, then \(\frac{4\sin \beta – 3\cos \beta}{4\sin \beta + 3\cos \beta} =\) [Term I CBSE 2021-2022]
(a) 0
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{3}{4}\)
View Answer
Solution:
4 tan β = 3 ⇒ tan β = 3/4
Divide numerator and denominator by cos β:
\(\frac{4\sin \beta – 3\cos \beta}{4\sin \beta + 3\cos \beta} = \frac{4\tan \beta – 3}{4\tan \beta + 3} = \frac{4 \times \frac{3}{4} – 3}{4 \times \frac{3}{4} + 3}\) \(= \frac{3 – 3}{3 + 3} = \frac{0}{6} = 0\)
Answer: (a) 0
12. If tan a + cot a = 2, then tan²⁰a + cot²⁰a = [Term I CBSE 2021-2022]
(a) 0
(b) 2
(c) 20
(d) 220
View Answer
Solution:
tan a + cot a = 2
Let tan a = x, then cot a = 1/x
x + 1/x = 2 ⇒ x² – 2x + 1 = 0 ⇒ (x – 1)² = 0 ⇒ x = 1
So tan a = 1, cot a = 1
tan²⁰a + cot²⁰a = 1²⁰ + 1²⁰ = 1 + 1 = 2
Answer: (b) 2
13.In the given figure, D is the mid-point of BC, then the value of \(\frac{\cot y^\circ}{\cot x^\circ}\) is [Term I CBSE 2021-2022]
(a) 2
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
View Answer
Solution:
Since D is the midpoint of BC, BD = DC
In right triangles ABD and ADC:
cot x° = adjacent/opposite = BD/AD
cot y° = DC/AD
Since BD = DC, cot x° = cot y°
So \(\frac{\cot y^\circ}{\cot x^\circ} = 1\)
Note: This doesn’t match any option, suggesting there might be additional information in the figure.
14. In ΔABC right angled at B, if tan A = √3, then cos A cos C – sin A sin C = [Term I CBSE 2021-2022]
(a) -1
(b) 0
(c) 1
(d) \(\frac{\sqrt{3}}{2}\)
View Answer
Solution:
tan A = √3 ⇒ A = 60°
Since triangle is right-angled at B, A + C = 90° ⇒ C = 30°
cos A cos C – sin A sin C = cos 60° cos 30° – sin 60° sin 30°
= (1/2)(√3/2) – (√3/2)(1/2) = √3/4 – √3/4 = 0
Answer: (b) 0
15. If the angles of ΔABC are in the ratio 1:1:2, then the value of \(\frac{\sec A}{\csc B} – \frac{\tan A}{\cot B}\) is [Term I CBSE 2021-2022]
(a) 0
(b) \(\frac{1}{2}\)
(c) 1
(d) \(\frac{\sqrt{3}}{2}\)
View Answer
Solution:
Angles ratio 1:1:2 ⇒ A = B = x, C = 2x
A + B + C = 180° ⇒ x + x + 2x = 180° ⇒ x = 45°
So A = B = 45°, C = 90°
\(\frac{\sec A}{\csc B} – \frac{\tan A}{\cot B} = \frac{\sec 45^\circ}{\csc 45^\circ} – \frac{\tan 45^\circ}{\cot 45^\circ}\) \(= \frac{\sqrt{2}}{\sqrt{2}} – \frac{1}{1} = 1 – 1 = 0\)
Answer: (a) 0
16.If 2 sin²β – cos²β = 2, then β is [Term I CBSE 2021-2022]
(a) 0°
(b) 90°
(c) 45°
(d) 30°
View Answer
Solution:2 sin²β – cos²β = 22(1 – cos²β) – cos²β = 2
2 – 2cos²β – cos²β = 2
-3cos²β = 0 ⇒ cos²β = 0 ⇒ β = 90°
Answer: (b) 90°
17. If 1 + sin²α = 3 sin α cos α, then values of cot α are [Term I CBSE 2021-2022]
(a) -1,1
(b) 0,1
(c) 1,2
(d) -1,-1
View Answer
Solution:
1 + sin²α = 3 sin α cos α
Divide both sides by sin²α:
1/sin²α + 1 = 3 cos α/sin α
csc²α + 1 = 3 cot α
1 + cot²α + 1 = 3 cot α (since csc²α = 1 + cot²α)
cot²α – 3 cot α + 2 = 0
(cot α – 1)(cot α – 2) = 0
cot α = 1 or 2
Answer: (c) 1,2
18. If x = 2 sin²θ and y = 2 cos²θ + 1, then find x + y [Term I CBSE 2021-2022]
View Answer
Solution:
x + y = 2 sin²θ + 2 cos²θ + 1 = 2(sin²θ + cos²θ) + 1 = 2(1) + 1 = 3
Answer: 3