Practice MCQ Questions and Answer on Trigonometry

11.

The value of 4sin²30° + 3cot²60° is:

• (A) 1
• (B) $$\frac{1}{{\sqrt 2 }}$$
• (C) 2
• (D) 0

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 $$\eqalign{ & 4{\sin ^2}{30^ \circ } + 3{\cot ^2}{60^ \circ } \cr & = 4 \times {\left( {\frac{1}{2}} \right)^2} + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = 4 \times \frac{1}{4} + 3 \times \frac{1}{3} \cr & = 1 + 1 \cr & = 2 \cr} $$

12.

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

Show Answer

 $$\eqalign{ & {\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{1 - \frac{{\cos \theta }}{{\sin \theta }}}}{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}} \right)^2} + 1 \cr & = {\left( {\frac{{\sin \theta - \cos \theta }}{{\sin \theta }} \times \frac{{\cos \theta }}{{\cos \theta - \sin \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{\sin \theta - \cos \theta \times \cos \theta }}{{ - \sin \theta \left( {\sin \theta - \cos \theta } \right)}}} \right)^2} + 1 \cr & = {\left( {\frac{{ - \cos \theta }}{{\sin \theta }}} \right)^2} + 1 \cr & = {\cot ^2}\theta + 1 \cr & = {\text{cose}}{{\text{c}}^2}\theta \cr} $$

13.

If tan(2θ + 45°) = cot3θ, where (2θ + 45°) and 3θ are acute angles, then the value of θ is?

• (A) 5°
• (B) 9°
• (C) 12°
• (D) 15°

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 $$\eqalign{ & {\text{tan}}\left( {2\theta + {{45}^ \circ }} \right) = \cot 3\theta \cr & \left[ {{\text{If tan A}} = {\text{cot B}}} \right] \cr & ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr & \Rightarrow \left( {2\theta + {{45}^ \circ }} \right) + 3\theta = {90^ \circ } \cr & \Rightarrow 5\theta + {45^ \circ } = {90^ \circ } \cr & \Rightarrow \theta = \frac{{45}}{5} \cr & \Rightarrow \theta = {9^ \circ } \cr} $$

14.

If secθ + tanθ = m(>1), then the value of sinθ is (0° θ 90°)

• (A) $$\frac{{1 - {m^2}}}{{1 + {m^2}}}$$
• (B) $$\frac{{{m^2} - 1}}{{{m^2} + 1}}$$
• (C) $$\frac{{{m^2} + 1}}{{{m^2} - 1}}$$
• (D) $$\frac{{1 + {m^2}}}{{1 - {m^2}}}$$

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 $$\eqalign{ & \sec \theta + \tan \theta = m\,.......(i) \cr & {\text{then, }}\sec \theta - \tan \theta = \frac{1}{m}\,......(ii) \cr & Because{\text{ }}{\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & {\text{From equation (i)}} - {\text{(ii)}} \cr & 2\tan \theta = m - \frac{1}{m} \cr} $$ $$\eqalign{ & {\text{tan}}\theta = \frac{{{m^2} - 1}}{{2m}} \cr & \sin \theta = \frac{{{m^2} - 1}}{{{m^2} + 1}} \cr} $$

15.

Simplify the following: $$\frac{{\cos x - \sqrt 3 \sin x}}{2}$$

• (A) $$\cos \left( {\frac{\pi }{3} - x} \right)$$
• (B) $$\sin \left( {\frac{\pi }{3} + x} \right)$$
• (C) $$\cos \left( {\frac{\pi }{3} + x} \right)$$
• (D) $$\sin \left( {\frac{\pi }{3} - x} \right)$$

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 $$\eqalign{ & \frac{{\cos x - \sqrt 3 \sin x}}{2} \cr & \Rightarrow \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x \cr & \Rightarrow \cos {60^ \circ }.\cos x - \sin {60^ \circ }.\sin x \cr & \Rightarrow \cos \left( {{{60}^ \circ } + x} \right) \cr & \Rightarrow \cos \left( {\frac{\pi }{3} + x} \right) \cr} $$

16.

If sin31° = $$\frac{x}{y}{\text{,}}$$ then the value of sec31° - sin59° is?

• (A) $$\frac{{{x^2}}}{{y\sqrt {{y^2} + {x^2}} }}$$
• (B) $$\frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }}$$
• (C) $$ - \frac{{{y^2}}}{{\sqrt {{y^2} - {x^2}} }}$$
• (D) $$ - \frac{{{x^2}}}{{\sqrt {{y^2} - {x^2}} }}$$

Show Answer

 sin31° = $$\frac{x}{y}{\text{,}}$$ ∴ sec31° - sin59° $$\eqalign{ & = \frac{y}{{\sqrt {{y^2} - {x^2}} }} - \frac{{\sqrt {{y^2} - {x^2}} }}{y} \cr & = \frac{{{y^2} - ({y^2} - {x^2})}}{{y\sqrt {{y^2} - {x^2}} }} \cr & = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$

17.

The value of $$\frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} = ?$$

• (A) cos60°
• (B) tan60°
• (C) sin60°
• (D) sin30°

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 $$\eqalign{ & \frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} \cr & = \frac{{2 \times \sqrt 3 }}{{1 + 3}} \cr & = \frac{{\sqrt 3 }}{2} \cr & = \sin {60^ \circ } \cr} $$

18.

If $$\frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a$$   and $$\frac{{{\text{sin }}\alpha }}{{{\text{sin }}\beta }} = b{\text{,}}$$   then the value of $${\sin ^2}\beta $$  in terms of a and b is?

• (A) $$\frac{{{a^2} + 1}}{{{a^2} - {b^2}}}$$
• (B) $$\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}$$
• (C) $$\frac{{{a^2} - 1}}{{{a^2} - {b^2}}}$$
• (D) $$\frac{{{a^2} - 1}}{{{a^2} + {b^2}}}$$

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 $$\eqalign{ & \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a{\text{ }} \cr & \Rightarrow \cos {\text{ }}\alpha = a{\text{ }}\cos {\text{ }}\beta \cr & {\text{On squaring both sides}} \cr & {\cos ^2}\alpha = {a^2}{\cos ^2}\beta \cr & \Rightarrow 1 - {\sin ^2}\alpha = {a^2}\left( {1 - {{\sin }^2}\beta } \right)....(i) \cr & {\text{Again, }}\sin \alpha = {\text{ }}b\sin \beta \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\sin ^2}\alpha = {\text{ }}{b^2}{\sin ^2}\beta \cr & {\text{Put the value of }}{\sin ^2}\alpha {\text{ in equation (i)}} \cr & \Rightarrow {\text{1}} - {b^2}{\sin ^2}\beta = {a^2} - {a^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = {a^2}si{n^2}\beta - {b^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = si{n^2}\beta \left( {{a^2} - {b^2}} \right) \cr & \Rightarrow si{n^2}\beta = \frac{{{a^2} - 1}}{{{a^2} - {b^2}}} \cr} $$

19.

If $$\frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = 3{\text{,}}$$    then the value of $${\sin ^4}\theta $$   is?

• (A) $$\frac{{16}}{{25}}$$
• (B) $$\frac{2}{5}$$
• (C) $$\frac{1}{5}$$
• (D) $$\frac{3}{5}$$

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 $$\eqalign{ & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & {\text{Find }}{\sin ^4}\theta = ? \cr & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & \left( {{\text{by C & D}}} \right) \cr & \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} = \frac{{3 + 1}}{{3 - 1}} \cr & \Rightarrow {\text{tan}}\theta = 2 \cr & {\text{tan}}\theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{1} \cr & \Rightarrow {\sin ^4}\theta \Rightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$

20.

If θ is positive acute angle and 7cos²ÃƒÂƒÃ‚ŽÃ‚¸ + 3sin²ÃƒÂƒÃ‚ŽÃ‚¸ = 4, then value of θ is?

• (A) 60°
• (B) 30°
• (C) 45°
• (D) 90°

Show Answer

 $$\eqalign{ & {\text{7co}}{{\text{s}}^2}\theta + 3{\sin ^2}\theta = 4 \cr & 7{\text{co}}{{\text{s}}^2}\theta + 3\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) - 4 = 0 \cr & 7{\text{co}}{{\text{s}}^2}\theta - 3{\text{co}}{{\text{s}}^2}\theta + 3 - 4 = 0 \cr & 4{\text{co}}{{\text{s}}^2}\theta = 1 \cr & {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & {\text{cos}}\theta = \frac{1}{2} \cr & \cos \theta = {\text{cos}}{60^ \circ } \cr & \theta = {60^ \circ } \cr} $$