Trigonometry MCQ Questions and Solutions with Explanations

21.
$${\left( {\frac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} + 1 = ?$$

A. cosec²θ

B. sec²θ

C. cos²θ

D. sin²θ

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22.
The value of $${\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1,$$ 0° < θ < 90°, is equal to:

A. cosec²θ

B. sin²θ

C. cos²θ

D. sec²θ

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23.
If sin(A + B) = cos(A + B), what is the value of tanA?

A. $$\frac{{1 - \tan B}}{{1 + \tan B}}$$

B. $$\frac{{1 + \tan B}}{{1 - \tan B}}$$

C. $$\frac{{1 + \sec B}}{{1 - \sec B}}$$

D. $$\frac{{1 - {\text{cosec}}\,B}}{{1 + {\text{cosec}}\,B}}$$

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24.
$$\frac{{{{\left( {1 + \cos \theta } \right)}^2} + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} = ?$$

A. cosθ(1 + sinθ)

B. secθ(1 + sinθ)

C. 2cosθ(1 + secθ)

D. 2secθ(1 + secθ)

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25.
If cos(A - B) = $$\frac{{\sqrt 3 }}{2}$$ and sec A = 2, 0° ≤ A ≤ 90°, 0° ≤ B ≤ 90° then what is the measure of B?

A. 60°

B. 0°

C. 30°

D. 90°

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26.
The value of $$\frac{{2{{\cos }^3}\theta - \cos \theta }}{{\sin \theta - 2{{\sin }^3}\theta }}:$$

A. secθ

B. sinθ

C. cotθ

D. tanθ

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27.
The value of $$\frac{{\sec \theta \left( {\sin \theta - 2{{\sin }^3}\theta } \right)}}{{2{{\cos }^3}\theta - \cos \theta }}$$ is:

A. tanθ.cosθ

B. secθ.sinθ

C. secθ.tanθ

D. sinθ.cosθ

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28.
What is the value of $$\frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}}?$$

A. cosx + sinx

B. sinx - cosx

C. secx + tanx

D. secx - tanx

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Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & \frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}} \cr & \Rightarrow \frac{{1 + 2{{\tan }^2}x - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{1 + 2\left( {{{\sec }^2}x - 1} \right) - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{2{{\sec }^2}x - 1 - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{2\sec x\left( {\sec x - \tan x} \right)}}{{\sec x - \tan x}} - \frac{1}{{\sec x - \tan x}} \cr & \Rightarrow 2\sec x - \frac{1}{{\sec x - \tan x}} \cr & \Rightarrow 2\sec x - \frac{1}{{\sec x - \tan x}} \times \frac{{\sec x + \tan x}}{{\sec x + \tan x}} \cr & \Rightarrow 2\sec x - \frac{{\sec x + \tan x}}{{{{\sec }^2}x - {{\tan }^2}x}} \cr & \Rightarrow 2\sec x - \sec x - \tan x \cr & \Rightarrow \sec x - \tan x \cr & \cr & {\bf{Alternative:}} \cr & \frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}} \cr & {\text{By putting }}x = {45^ \circ }{\text{ in equation}} \cr & \Rightarrow \frac{{1 + 2{{\tan }^2}{{45}^ \circ } - 2\sec {{45}^ \circ }\tan {{45}^ \circ }}}{{\sec {{45}^ \circ } - \tan {{45}^ \circ }}} \cr & \Rightarrow \frac{{1 + 2 - 2\sqrt 2 }}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{{3 - 2\sqrt 2 }}{{\sqrt 2 - 1}} \times \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \cr & \Rightarrow 3\sqrt 2 - 2\sqrt 2 - 1 \cr & \Rightarrow \sqrt 2 \cr & {\text{By satisfying in options}} \cr & \Rightarrow \sec x - \tan x \Rightarrow \sqrt 2 - 1 \cr} $$

29.
If tan40° = α, then find $$\frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ } \cdot \tan {{310}^ \circ }}}$$

A. $$\frac{{1 - {\alpha ^2}}}{\alpha }$$

B. $$\frac{{1 + {\alpha ^2}}}{{2\alpha }}$$

C. $$\frac{{1 - {\alpha ^2}}}{{2\alpha }}$$

D. $$\frac{{1 + {\alpha ^2}}}{\alpha }$$

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30.
If $$\frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c}$$ and A = 60°, x = 15°, then the value of $${\left( {\frac{{a + c}}{b}} \right)^2} + {\left( {\frac{{a - c}}{b}} \right)^2}$$ is . . . . . . . .

A. 0

B. 4

C. 2

D. 1

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Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & A = {60^ \circ },\,\,x = {15^ \circ }{\text{ given,}} \cr & \Rightarrow \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{60}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}} = \frac{a}{b} \cr & \Rightarrow \frac{{\cos {{75}^ \circ }}}{{\cos {{135}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{\sin {{15}^ \circ }}}{{ - \sin {{45}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{{{{\sin }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = 2{\sin ^2}{15^ \circ }........\left( {\text{i}} \right) \cr & \Rightarrow \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{180}^ \circ }} \right)}} = \frac{b}{c} \cr & \Rightarrow \frac{{ - \sin {{45}^ \circ }}}{{ - \cos {{15}^ \circ }}} = \frac{b}{c} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = \frac{{{{\cos }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = 2{\cos ^2}{15^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{Now, }}{\left( {\frac{a}{b} + \frac{c}{b}} \right)^2} + {\left( {\frac{a}{b} - \frac{c}{b}} \right)^2} \cr & = 2\left[ {\frac{{{a^2}}}{{{b^2}}} + \frac{{{c^2}}}{{{b^2}}}} \right] \cr & = 2\left[ {2{{\sin }^2}{{15}^ \circ } + 2{{\cos }^2}{{15}^ \circ }} \right] \cr & = 4 \cr} $$

Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

21. $${\left( {\frac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} + 1 = ?$$

Answer & Solution

22. The value of $${\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1,$$ 0° < θ < 90°, is equal to:

Answer & Solution

23. If sin(A + B) = cos(A + B), what is the value of tanA?

Answer & Solution

24. $$\frac{{{{\left( {1 + \cos \theta } \right)}^2} + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} = ?$$

Answer & Solution

25. If cos(A - B) = $$\frac{{\sqrt 3 }}{2}$$ and sec A = 2, 0° ≤ A ≤ 90°, 0° ≤ B ≤ 90° then what is the measure of B?

Answer & Solution

26. The value of $$\frac{{2{{\cos }^3}\theta - \cos \theta }}{{\sin \theta - 2{{\sin }^3}\theta }}:$$

Answer & Solution

27. The value of $$\frac{{\sec \theta \left( {\sin \theta - 2{{\sin }^3}\theta } \right)}}{{2{{\cos }^3}\theta - \cos \theta }}$$ is:

Answer & Solution

28. What is the value of $$\frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}}?$$

Answer & Solution

29. If tan40° = α, then find $$\frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ } \cdot \tan {{310}^ \circ }}}$$

Answer & Solution

30. If $$\frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c}$$ and A = 60°, x = 15°, then the value of $${\left( {\frac{{a + c}}{b}} \right)^2} + {\left( {\frac{{a - c}}{b}} \right)^2}$$ is . . . . . . . .

Answer & Solution

Read More Section(Trigonometry)

21.
$${\left( {\frac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} + 1 = ?$$

22.
The value of $${\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1,$$ 0° < θ < 90°, is equal to:

23.
If sin(A + B) = cos(A + B), what is the value of tanA?

24.
$$\frac{{{{\left( {1 + \cos \theta } \right)}^2} + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} = ?$$

25.
If cos(A - B) = $$\frac{{\sqrt 3 }}{2}$$ and sec A = 2, 0° ≤ A ≤ 90°, 0° ≤ B ≤ 90° then what is the measure of B?

26.
The value of $$\frac{{2{{\cos }^3}\theta - \cos \theta }}{{\sin \theta - 2{{\sin }^3}\theta }}:$$

27.
The value of $$\frac{{\sec \theta \left( {\sin \theta - 2{{\sin }^3}\theta } \right)}}{{2{{\cos }^3}\theta - \cos \theta }}$$ is:

28.
What is the value of $$\frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}}?$$

29.
If tan40° = α, then find $$\frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ } \cdot \tan {{310}^ \circ }}}$$

30.
If $$\frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c}$$ and A = 60°, x = 15°, then the value of $${\left( {\frac{{a + c}}{b}} \right)^2} + {\left( {\frac{{a - c}}{b}} \right)^2}$$ is . . . . . . . .