Practice MCQ Questions and Answer on Trigonometry

81.

If 2cos²ÃƒÂƒÃ‚ŽÃ‚¸ + 3sinθ = 3, where 0° θ 90°, then what is the value of sin²2θ + cos²ÃƒÂƒÃ‚ŽÃ‚¸ + tan²2θ + cosec²2θ?

• (A) $$\frac{{29}}{6}$$
• (B) $$\frac{{29}}{3}$$
• (C) $$\frac{{35}}{6}$$
• (D) $$\frac{{35}}{{12}}$$

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 $$\eqalign{ & 2{\cos ^2}\theta + 3\sin \theta = 3 \cr & {\text{Let }}\theta = {30^ \circ } \cr & 2{\cos ^2}{30^ \circ } + 3\sin {30^ \circ } = 3 \cr & 2 \times \frac{3}{4} + 3 \times \frac{1}{2} = 3 \cr & 3 = 3 \cr & {\sin ^2}2\theta + {\cos ^2}\theta + {\tan ^2}2\theta + {\text{cose}}{{\text{c}}^2}2\theta \cr & = {\sin ^2}{60^ \circ } + {\cos ^2}{30^ \circ } + {\tan ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{60^ \circ } \cr & = \frac{3}{4} + \frac{3}{4} + 3 + \frac{4}{3} \cr & = \frac{3}{2} + 3 + \frac{4}{3} \cr & = \frac{{9 + 18 + 8}}{6} \cr & = \frac{{35}}{6} \cr} $$

82.

If 1 + sin²ÃƒÂƒÃ‚ŽÃ‚¸ - 3sinθcosθ = 0, then the value of cotθ is:

• (A) 2
• (B) $$\frac{1}{3}$$
• (C) $$\frac{1}{2}$$
• (D) 0

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 1 + sin2θ - 3sinθ.cosθ = 0 ⇒ 1 - 2sinθ.cosθ = sinθ.cosθ - sin2θ ⇒ sin2θ + cos2θ - 2sinθ.cosθ = sinθ(cosθ - sinθ) ⇒ (cosθ - sinθ)2 = sinθ(cosθ - sinθ) ⇒ cosθ - sinθ = sinθ ⇒ cosθ = 2sinθ ⇒ cotθ = 2

83.

If tan²A - 6tanA + 9 = 0, 0° A 90°, What is the value of 6cotA + $$8\sqrt {10} $$ cosA?

• (A) $$10\sqrt {10} $$
• (B) 14
• (C) 10
• (D) $$8\sqrt {10} $$

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 tan2A - 6tanA + 9 = 0 tan2A - 3tanA - 3tanA + 9 = 0 tanA(tanA - 3) - 3(tanA - 3) = 0 (tanA - 3)(tanA - 3) = 0 tanA $$ = \frac{3}{1} = \frac{P}{B}$$ H2 = P2 + B2 $$\eqalign{ & H = \sqrt {9 + 1} = \sqrt {10} \cr & 6\cot A + 8\sqrt {10} \cos A \cr & = 6\left( {\frac{B}{P}} \right) + 8\sqrt {10} \left( {\frac{B}{H}} \right) \cr & = 6\left( {\frac{1}{3}} \right) + 8\sqrt {10} \left( {\frac{1}{{\sqrt {10} }}} \right) \cr & = 2 + 8 \cr & = 10 \cr} $$

84.

In ΔABC, âˆÂ B = 90° and AB : BC = 2 : 1, then value of (sinA + cotC) = ?

• (A) $$3 + \sqrt 5 $$
• (B) $$\frac{{2 + \sqrt 5 }}{{2\sqrt 5 }}$$
• (C) $$2 + \sqrt 5 $$
• (D) $$3\sqrt 5 $$

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 $$\eqalign{ & {\text{AC}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 \cr & {\text{sin A}} + \operatorname{cotC} \cr & \frac{{{\text{BC}}}}{{{\text{AC}}}} + \frac{{{\text{BC}}}}{{{\text{AB}}}} \cr & \frac{1}{{\sqrt 5 }} + \frac{1}{2} \cr & \Rightarrow \frac{{2 + \sqrt 5 }}{{2\sqrt 5 }} \cr} $$

85.

The value of $$\frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}}$$        is equal to:

• (A) 2cosθ
• (B) cosecθsecθ
• (C) 2sinθ
• (D) sinθcosθ

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 $$\eqalign{ & \frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}} \cr & = \frac{{\left( {\sec \theta - \tan \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {\frac{{1 + \tan \theta }}{{\tan \theta }}} \right)}} \cr & = \frac{{\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {\sin \theta + \cos \theta } \right)}}{{\left( {1 + \tan \theta } \right)\left( {\sin \theta + \frac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right)}} \cr & = \frac{{1\left( {\sin \theta + \cos \theta } \right)}}{{\left( {1 + \tan \theta } \right)\left( {\frac{1}{{\sin \theta }}} \right)}} \cr & = \frac{{\left( {\sin \theta + \cos \theta } \right)}}{{\left( {\cos \theta + \sin \theta } \right)}}\sin \theta .\cos \theta \cr & = \sin \theta .\cos \theta \cr} $$

86.

$${\left( {\frac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3} - \cos \theta }}} \right)^2} + 1,$$     θ â‰Â  45° is equal to:

• (A) cosec2θ
• (B) sec2θ
• (C) cot2θ
• (D) 2tan2θ

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 $$\eqalign{ & {\left( {\frac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3} - \cos \theta }}} \right)^2} + 1 \cr & = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}{\left( {\frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)^2} + 1 \cr & = {\tan ^2}\theta + 1 \cr & = {\sec ^2}\theta \cr} $$

87.

The value of coses²60° + sec²60° - cot²60° + tan²30° will be?

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

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 $$\eqalign{ & {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\text{se}}{{\text{c}}^2}{60^ \circ } - {\text{co}}{{\text{t}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{30^ \circ } \cr & = {\left( {\frac{2}{{\sqrt 3 }}} \right)^2} + {\left( 2 \right)^2} - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = \frac{4}{3} + 4 - \frac{1}{3} + \frac{1}{3} \cr & = \frac{{16}}{3} \cr & = 5\frac{1}{3} \cr} $$

88.

The value of expression cos²45° + cos²135° + cos²225° + cos²315° is:

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

Show Answer

 cos245° + cos2135° + cos2225° + cos2315° = (cos245° + cos2315°) + (cos2135° + cos2225°)   [Here cos2α + cos2β = 1, where α + β = 360°] = 1 + 1 = 2

89.

Evaluate the following expression in terms of trigonometric ratios.
$$\frac{{{{\cot }^2}A\left( {\sec A - 1} \right)}}{{1 + \sin A}}$$

• (A) $$\frac{{{{\sec }^2}A\left( {1 + \sin A} \right)}}{{1 - \sec A}}$$
• (B) $$\frac{{{{\sec }^2}A\left( {1 - \sin A} \right)}}{{1 + \sec A}}$$
• (C) $$\frac{{{{\cot }^2}A\left( {1 + \sin A} \right)}}{{1 - \sec A}}$$
• (D) $$\frac{{{{\cot }^2}A\left( {1 - \sin A} \right)}}{{1 - \sec A}}$$

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 $$\eqalign{ & \frac{{{{\cot }^2}A\left( {\sec A - 1} \right)}}{{\left( {1 + \sin A} \right)}} \cr & = \frac{{{{\cos }^2}A\left( {1 - \cos A} \right)}}{{{{\sin }^2}A\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\left( {1 - {{\sin }^2}A} \right)}}{{\left( {1 - {{\cos }^2}A} \right)}} \times \frac{{\left( {1 - \cos A} \right)}}{{\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right).\left( {1 - \cos A} \right)}}{{\left( {1 + \cos A} \right)\left( {1 - \cos A} \right).\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\frac{{\sec A\left( {1 - \sin A} \right)}}{{\left( {\sec A + 1} \right)}}}}{{\sec A}} \cr & = \frac{{{{\sec }^2}A\left( {1 - \sin A} \right)}}{{\left( {1 + \sec A} \right)}} \cr} $$

90.

If $$\sin \alpha + \cos \beta = 2;$$   $$\left( {{0^ \circ } \leqslant \beta \alpha \leqslant {{90}^ \circ }} \right){\text{,}}$$     then $${\text{sin}}\,{\left( {\frac{{2\alpha + \beta }}{3}} \right)^ \circ }$$   is?

• (A) $${\text{sin }}\frac{\alpha }{2}$$
• (B) $$\cos \frac{\alpha }{3}$$
• (C) $${\text{sin }}\frac{\alpha }{3}$$
• (D) $$\cos \frac{{2\alpha }}{3}$$

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 $$\eqalign{ & \sin {\text{ }}\alpha + \cos \beta = 2 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Leftrightarrow {\text{sin }}{90^ \circ } + {\text{cos }}{0^ \circ } = 2 \cr & \Leftrightarrow 1 + 1 = 2 \cr & \Leftrightarrow 2 = 2\left[ {{\text{Matched}}} \right] \cr & {\text{So, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2\alpha + \beta }}{3}} \right)^ \circ }{\text{ }} \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2 \times {{90}^ \circ } + {0^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{{{180}^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{60^ \circ } = \cos {30^ \circ } = \frac{{\sqrt 3 }}{2} \cr & {\text{Take cos}}\frac{\alpha }{3} = \cos \frac{{{{90}^ \circ }}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{cos 3}}{0^ \circ } \cr & {\text{So, this is answer }}{\text{.}} \cr} $$