SSC CGL Trigonometry Questions with Detailed Solutions

Given:
Trigonometric identity: $$1 + \cot^2\theta = \csc^2\theta$$​​
Formula Used:
​$$\cosec^2\theta - \cot^2\theta = 1$$​​
Solution:
Rearranging the given identity gives $$\cosec^2\theta - \cot^2\theta$$​ = 1.

This identity holds true for all values of $$\theta$$ where the functions are defined.
Therefore, for $$\theta$$​ = 60°

$$\cosec^2(60°) - \cot^2(60°)$$=  1.
So the correct answer is (a).

​$$\textbf{Given:}\cos \theta = -\frac{4}{5}\\ \ \\\theta \text{ lies in the third quadrant} \\ \ \\\text{Find } (\sin \theta + \cos \theta)^2\\ \ \\\textbf{Concept Used:}\\ \ \\\sin^2\theta + \cos^2\theta = 1\\ \ \\(a+b)^2 = a^2 + b^2 + 2ab\\ \ \\\text{In III quadrant: } \sin\theta < 0,\; \cos\theta < 0\\ \ \\\textbf{Formula Used:}\\ \ \\\sin^2\theta = 1 - \cos^2\theta\\ \ \\(\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta \\ \ \\\textbf{Solution:}\\ \ \\\cos\theta = -\frac{4}{5} \\ \ \\\sin^2\theta = 1 - \cos^2\theta \\ \ \\= 1 - \left(\frac{4}{5}\right)^2= 1 - \frac{16}{25}\\ \ \\= \frac{9}{25}\\ \ \\\sin\theta = -\frac{3}{5}\\ \ \\(\sin\theta + \cos\theta)^2\\ \ \\= \left(-\frac{3}{5} - \frac{4}{5}\right)^2\\ \ \\= \left(-\frac{7}{5}\right)^2\\ \ \\= \frac{49}{25}\\ \ \\\frac{49}{25}\\ \ \\$$​​

Given:
​$$\sin A + \cos A = \frac{\sqrt{3}}{2}$$​​
Formula Used:
​$$(a + b)^2 = a^2 + b^2 + 2ab$$​​
Solution:
The expression we need to find is $$\sin^2 A + \cos^2 A + 2 \sin A \cos A$$​​
By algebraic identity, this expression is exactly equal to $$(\sin A + \cos A)^2$$​​
Substitute the given value of $$(\sin A + \cos A)$$​ into the expression:
​$$(\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$$​​
Final Answer
So the correct answer is (c)

Given:
​$$\cot A = 3$$​​
Angle A is an acute angle.
Formula Used:
​$$\cot A = \frac{\text{Base}}{\text{Perpendicular}}$$​​
​$$\sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$$​​
​$$\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2$$​​
Solution:
We can write $$\cot A = \frac{3}{1}$$​. Thus, Base = 3 and Perpendicular = 1.
Using Pythagoras theorem:
​$$\text{Hypotenuse} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$$​​
​$$\sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{\sqrt{10}}$$​​
Rationalizing the denominator:
​$$\sin A = \frac{1}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{10}}{10}$$​​
Final Answer
So the correct answer is (a)

Given:
Angle in radians = $$\frac{3\pi}{4}$$​​
Formula Used:
​$$\text{Degrees} = \text{Radians} \times \frac{180^\circ}{\pi}$$​​
Solution:
Angle in degrees = $$\frac{3\pi}{4} \times \frac{180^\circ}{\pi}$$​​
​$$= \frac{3 \times 180^\circ}{4}$$​​
​$$= 3 \times 45^\circ$$​​
​$$= 135^\circ$$​​
Final Answer
So the correct answer is (d)

Given:
​$$\sin x = \cos(2x - 10^\circ)$$​​
Formula Used:
If $$\sin A = \cos B, then A + B = 90^\circ$$​ (for acute angles).
Solution:
Apply the complementary angle property:
​$$x + (2x - 10^\circ) = 90^\circ\\ \ \\3x - 10^\circ = 90^\circ\\ \ \\3x = 100^\circ\\ \ \\x = \frac{100^\circ}{3} = 33.33^\circ$$​​
So the correct answer is (b).

Given:
A + B = $$90^\circ$$​​
​$$\tan A = 2$$​​
Formula Used:
​$$\tan(90^\circ - \theta) = \cot \theta$$​​
​$$\tan \theta \times \cot \theta = 1$$​​
Solution:
Since A + B = $$90^\circ$$​, we have B = $$90^\circ - A.$$​​
Therefore, $$\tan B = \tan(90^\circ - A) = \cot A.$$​​
Expression is $$\tan A \times \tan B.$$​​
Substituting the value, we get $$\tan A \times \cot A = 1$$​​
Final Answer
So the correct answer is (b)

Given:
​$$\sin x + \cos x = \sqrt{1.5}$$​​

Formula Used
​$$(\sin x + \cos x)^2 + (\sin x - \cos x)^2 = 2(\sin^2 x + \cos^2 x) = 2$$​​

Solution:
Let the required expression $$\sin x - \cos x = y$$​​
Square both the given and the target equations:
​$$(\sin x + \cos x)^2 = 1.5\\ \ \\(\sin x - \cos x)^2 = y^2\\ \ \\(\sin^2 x + \cos^2 x + 2\sin x\cos x) + (\sin^2 x + \cos^2 x - 2\sin x\cos x) = 1.5 + y^2$$​​
Simplify using the trigonometric identity $$\sin^2 x + \cos^2 x = 1$$​:
1 + 1 = 1.5 + $$y^2$$​​
2 = 1.5 + $$y^2$$​​
$$y^2$$​= 2 - 1.5 = 0.5
​$$y = \sqrt{0.5} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$​​
So the correct answer is (a)

Given:
​$$\tan A = \frac{1}{\sqrt{3}}$$​​
Formula Used:
​$$\tan 30^\circ = \frac{1}{\sqrt{3}}\\ \ \\\sin 30^\circ = \frac{1}{2}\\ \ \\\cos 30^\circ = \frac{\sqrt{3}}{2}$$​​
Solution: 
Identify the angle A from the given tangent value:
A = 30$$^\circ$$​​
​$$=(1 - \sin 30^\circ)(1 + \cos 30^\circ)\\ \ \\=(1 - \frac{1}{2})(1 + \frac{\sqrt{3}}{2})\\ \ \\=(\frac{1}{2})(\frac{2 + \sqrt{3}}{2})\\ \ \\=\frac{2 + \sqrt{3}}{4}$$​​
So the correct answer is (a)

Given
​$$\tan \theta + \cot \theta = 2$$​​
Solution:
If $$\tan \theta + \frac{1}{\tan \theta} = 2$$​,

then $$\tan \theta = 1$$​​
Thus, $$\theta = 45^\circ$$​​
​$$\sec 45^\circ + \cosec 45^\circ = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$$​​
answer (a)

Given:
​$$\sin \alpha = 4/5$$​​
Formula Used:
​$$\cot \alpha = \frac {Base} {Perp}$$​​

Solution:
P = 4, H = 5 $$\Rightarrow$$​B = 3.
​$$\cot \alpha = \frac{3}{4}$$​ 
$$\frac{1 - 3/4}{1 + 3/4}$$​ 

=$$\frac{1/4}{7/4}$$​​

= 1/7
So the correct answer is (d)

Given:
cosA = 4/5
sinB = 5/13
A, B ∈ (0, π/2)
Find cos(A − B)
Concept Used:
Trigonometric identity of cosine difference and Pythagoras identity.
Formula Used:
cos(A − B) = cosA cosB + sinA sinB
sin²θ + cos²θ = 1
Solution:
​$$\cos A = \frac{4}{5}\\ \ \\\sin A = \sqrt{1 - \cos^2 A}\\ \ \\= \sqrt{1 - \frac{16}{25}}\\ \ \\ = \sqrt{\frac{9}{25}}\\ \ \\= \frac{3}{5}\\ \ \\\sin B = \frac{5}{13}\\ \ \\\cos B = \sqrt{1 - \sin^2 B}\\ \ \\= \sqrt{1 - \frac{25}{169}}\\ \ \\= \sqrt{\frac{144}{169}}\\ \ \\= \frac{12}{13}\\ \ \\\cos(A - B) = \cos A \cos B + \sin A \sin B\\ \ \\= \left(\frac{4}{5} \times \frac{12}{13}\right) + \left(\frac{3}{5} \times \frac{5}{13}\right)\\ \ \\= \frac{48}{65} + \frac{15}{65}= \frac{63}{65}$$

​​

Given:
​$$\tan \theta = p - 1/p.$$​​
Formula Used:
​$$\sec^2 \theta = 1 + \tan^2 \theta$$​​
Solution:
​$$\tan^2 \theta = (p - 1/p)^2 = p^2 + 1/p^2 - 2$$​​
​$$\sec^2 \theta = 1 + (p^2 + 1/p^2 - 2)$$​​
​$$= p^2 + 1/p^2 - 1$$​​
Final Answer
So the correct answer is (b)