trigonometry — BSSC Quantitative Aptitude

50 Questions • 30 Minutes • Quantitative Aptitude Mock Test in Hindi and English

Sample Questions from this Test

Question 1:

If sec⁡A=135\sec A = \frac{13}{5} and AA is an acute angle, find the value of tan⁡A+sin⁡Atan⁡A−sin⁡A\frac{\tan A + \sin A}{\tan A - \sin A}.
A.9/4
B.4/9
C.25/16
D.16/9

1. sec⁡A=135  ⟹  cos⁡A=513\sec A = \frac{13}{5} \implies \cos A = \frac{5}{13}.
2. Using Pythagorean triplet (5,12,13)(5, 12, 13), Perpendicular P=12P = 12.
3. Divide numerator and denominator by sin⁡A\sin A:
tan⁡A+sin⁡Atan⁡A−sin⁡A=sec⁡A+1sec⁡A−1=135+1135−1=18/58/5=94\frac{\tan A + \sin A}{\tan A - \sin A} = \frac{\sec A + 1}{\sec A - 1} = \frac{\frac{13}{5} + 1}{\frac{13}{5} - 1} = \frac{18/5}{8/5} = \frac{9}{4}

Question 2:

If sin⁡θ=2xyx2+y2\sin \theta = \frac{2xy}{x^2 + y^2} and θ\theta is an acute angle, what is tan⁡θ\tan \theta in terms of xx and yy?
A.(2xy)/(x² - y²)
B.(x² - y²)/(2xy)
C.(x² + y²)/(x² - y²)
D.(x² - y²)/(x² + y²)

1. Perpendicular P=2xyP = 2xy and Hypotenuse H=x2+y2H = x^2 + y^2.
2. Base B=H2−P2=(x2+y2)2−(2xy)2=(x2−y2)2=x2−y2B = \sqrt{H^2 - P^2} = \sqrt{(x^2+y^2)^2 - (2xy)^2} = \sqrt{(x^2-y^2)^2} = x^2 - y^2.
3. tan⁡θ=PB=2xyx2−y2\tan \theta = \frac{P}{B} = \frac{2xy}{x^2 - y^2}.

Question 3:

If sin⁡A+cos⁡A=2\sin A + \cos A = \sqrt{2}, then find the value of sin⁡4A+cos⁡4A\sin^4 A + \cos^4 A.
A.1
B.1/2
C.3/4
D.5/8

1. Squaring both sides: (sin⁡A+cos⁡A)2=2  ⟹  1+2sin⁡Acos⁡A=2  ⟹  sin⁡Acos⁡A=12(\sin A + \cos A)^2 = 2 \implies 1 + 2\sin A\cos A = 2 \implies \sin A\cos A = \frac{1}{2}.
2. sin⁡4A+cos⁡4A=(sin⁡2A+cos⁡2A)2−2sin⁡2Acos⁡2A=1−2(12)2=1−12=12\sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2\sin^2 A\cos^2 A = 1 - 2\left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2}.

Question 4:

If sec⁡θ+tan⁡θ=5\sec \theta + \tan \theta = 5 and θ\theta is an acute angle, find the value of sin⁡θ\sin \theta.
A.12/13
B.5/13
C.24/25
D.7/25

1. Since sec⁡2θ−tan⁡2θ=1  ⟹  sec⁡θ−tan⁡θ=15\sec^2 \theta - \tan^2 \theta = 1 \implies \sec \theta - \tan \theta = \frac{1}{5}.
2. Adding: 2sec⁡θ=5+15=265  ⟹  sec⁡θ=135  ⟹  cos⁡θ=5132\sec \theta = 5 + \frac{1}{5} = \frac{26}{5} \implies \sec \theta = \frac{13}{5} \implies \cos \theta = \frac{5}{13}.
3. Subtracting: 2tan⁡θ=245  ⟹  tan⁡θ=1252\tan \theta = \frac{24}{5} \implies \tan \theta = \frac{12}{5}.
4. sin⁡θ=tan⁡θsec⁡θ=12/513/5=1213\sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{12/5}{13/5} = \frac{12}{13}.

Question 5:

If 3sin⁡θ+5cos⁡θ=53\sin \theta + 5\cos \theta = 5, find the positive value of 3cos⁡θ−5sin⁡θ3\cos \theta - 5\sin \theta.
A.3
B.5
C.√3
D.0

1. Identity: (asin⁡X+bcos⁡X)2+(acos⁡X−bsin⁡X)2=a2+b2(a\sin X + b\cos X)^2 + (a\cos X - b\sin X)^2 = a^2 + b^2.
2. Here a=3a = 3, b=5b = 5, and 3sin⁡θ+5cos⁡θ=53\sin \theta + 5\cos \theta = 5.
3. 52+(3cos⁡θ−5sin⁡θ)2=32+52  ⟹  (3cos⁡θ−5sin⁡θ)2=95^2 + (3\cos \theta - 5\sin \theta)^2 = 3^2 + 5^2 \implies (3\cos \theta - 5\sin \theta)^2 = 9.
4. 3cos⁡θ−5sin⁡θ=33\cos \theta - 5\sin \theta = 3 (positive value).

Question 6:

If sin⁡(2x+15∘)=cos⁡(x−5∘)\sin(2x + 15^\circ) = \cos(x - 5^\circ), where both are acute angles, find the value of xx.
A.26.67°
B.30°
C.25°
D.45°

1. Complementary relation: sin⁡A=cos⁡B  ⟹  A+B=90∘\sin A = \cos B \implies A + B = 90^\circ.
2. (2x+15∘)+(x−5∘)=90∘(2x + 15^\circ) + (x - 5^\circ) = 90^\circ.
3. 3x+10∘=90∘  ⟹  3x=80∘  ⟹  x=26.67∘3x + 10^\circ = 90^\circ \implies 3x = 80^\circ \implies x = 26.67^\circ.

Question 7:

Evaluate tan⁡10∘⋅tan⁡20∘⋅tan⁡30∘⋅tan⁡70∘⋅tan⁡80∘\tan 10^\circ \cdot \tan 20^\circ \cdot \tan 30^\circ \cdot \tan 70^\circ \cdot \tan 80^\circ.
A.1
B.√3
C.1/√3
D.1/2

1. tan⁡80∘=cot⁡10∘\tan 80^\circ = \cot 10^\circ and tan⁡70∘=cot⁡20∘\tan 70^\circ = \cot 20^\circ.
2. (tan⁡10∘cot⁡10∘)⋅(tan⁡20∘cot⁡20∘)⋅tan⁡30∘(\tan 10^\circ \cot 10^\circ) \cdot (\tan 20^\circ \cot 20^\circ) \cdot \tan 30^\circ.
3. 1⋅1⋅tan⁡30∘=131 \cdot 1 \cdot \tan 30^\circ = \frac{1}{\sqrt{3}}.

Question 8:

Evaluate the value of 5cos⁡260∘+4sec⁡230∘−tan⁡245∘sin⁡230∘+cos⁡230∘\frac{5\cos^2 60^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}.
A.67/12
B.37/12
C.57/12
D.27/5

1. Denominator = sin⁡230∘+cos⁡230∘=1\sin^2 30^\circ + \cos^2 30^\circ = 1.
2. Numerator = 5(1/2)2+4(2/3)2−12=54+163−1=15+64−1212=67125(1/2)^2 + 4(2/\sqrt{3})^2 - 1^2 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12}.

Question 9:

In ΔABC\Delta ABC, right-angled at CC, if sin⁡A=35\sin A = \frac{3}{5}, then find the value of cos⁡B\cos B.
A.3/5
B.4/5
C.1
D.2/5

1. In right-angled triangle ΔABC\Delta ABC at CC, A+B=90∘  ⟹  B=90∘−AA + B = 90^\circ \implies B = 90^\circ - A.
2. cos⁡B=cos⁡(90∘−A)=sin⁡A\cos B = \cos(90^\circ - A) = \sin A.
3. Given sin⁡A=35\sin A = \frac{3}{5}, therefore cos⁡B=35\cos B = \frac{3}{5}.

Question 10:

If tan⁡θ+cot⁡θ=2\tan \theta + \cot \theta = 2 where θ\theta is an acute angle, find the value of tan⁡3θ+cot⁡3θ+2tan⁡5θcot⁡7θ\tan^3 \theta + \cot^3 \theta + 2\tan^5 \theta \cot^7 \theta.
A.2
B.4
C.6
D.3

1. tan⁡θ+cot⁡θ=2  ⟹  θ=45∘\tan\theta + \cot\theta = 2 \implies \theta = 45^\circ (since tan⁡45∘=1\tan 45^\circ = 1 and cot⁡45∘=1\cot 45^\circ = 1).
2. Substituting tan⁡θ=1\tan\theta = 1 and cot⁡θ=1\cot\theta = 1:
13+13+2(15)(17)=1+1+2=41^3 + 1^3 + 2(1^5)(1^7) = 1 + 1 + 2 = 4

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