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Basics of Trigonometry (Level - 2)
Q1–10 of 50
1

Find the exact value of tan(15Β°).

Correct Answer: B. 2 βˆ’ √3
Explanation:

Solution: Express 15Β° as (45Β°βˆ’30Β°) and use tan(Aβˆ’B) formula. tan15Β° = (√3βˆ’1)/(√3+1) = 2βˆ’βˆš3


2

Find the exact value of cos(22.5Β°).

Correct Answer: B. $\frac{\sqrt{2 + \sqrt{2}}}{2}$
Explanation:

Use the cosine half-angle formula $\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos\theta}{2}}$ with $\theta = 45^\circ$.

$\cos(22.5^\circ) = \sqrt{\frac{1 + \sqrt{2}/2}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$.

3

Evaluate sin75Β°cos15Β° βˆ’ cos75Β°sin15Β°.


Correct Answer: C. √3/2
Explanation: No explanation available.
4

Evaluate tan(22.5Β°) + cot(22.5Β°).

Correct Answer: B. 2√2
Explanation:

Solution: tanθ+cotθ = 2/sin2θ. For θ=22.5°: 2/sin45° = 2√2.


5

Find the exact value of sin(15Β°) + cos(15Β°).

Correct Answer: A. √6/2
Explanation:

Solution: sin15Β°=(√6βˆ’βˆš2)/4, cos15Β°=(√6+√2)/4. Sum = √6/2.


6

Evaluate the expression:

$$\frac{\tan(75^\circ) - \tan(15^\circ)}{1 + \tan(75^\circ)\tan(15^\circ)}$$
Correct Answer: B. √3
Explanation:

Solution: This is tan(Aβˆ’B) with A=75Β°,B=15Β°: tan60Β° = √3.


7

Find the exact value of tan(67.5Β°) βˆ’ tan(22.5Β°).

Correct Answer: C. 2
Explanation:

Solution: tan67.5Β° = √2+1, tan22.5Β° = √2βˆ’1. Difference = 2


8

Evaluate 8sin15Β°.cos15Β°.cos30Β°.


Correct Answer: B. √3
Explanation: No explanation available.
9

Find the value of:

$$\frac{1 - \tan^2(22.5^\circ)}{1 + \tan^2(22.5^\circ)}$$
Correct Answer: C. √2/2
Explanation:

Solution: This equals cos(2·22.5°) = cos45° = √2/2.


10

Evaluate cos(22.5Β°).cos(67.5Β°).

Correct Answer: B. √2/4
Explanation:

Solution: cos67.5°=sin22.5°, so expression = sin22.5°cos22.5° = sin45°/2 = √2/4.


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