Practice MCQ Questions and Answer on Trigonometry

341.

If xtan60° + cos45° = sec45°, then the value of x² + 1 is?

• (A) $$\frac{6}{7}$$
• (B) $$\frac{7}{6}$$
• (C) $$\frac{5}{6}$$
• (D) $$\frac{6}{5}$$

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 $$\eqalign{ & \Rightarrow x\tan {60^ \circ } + \cos {45^ \circ } = \sec {45^ \circ } \cr & \Rightarrow x.\sqrt 3 + \frac{1}{{\sqrt 2 }} = \sqrt 2 \cr & \Rightarrow x\sqrt 6 + 1 = 2 \cr & \Rightarrow x = \frac{1}{{\sqrt 6 }} \cr }$$ Squaring both sides and addition 1 both sides $$\eqalign{ & \Rightarrow {x^2} + 1 = {\left( {\frac{1}{{\sqrt 6 }}} \right)^2} + 1 \cr & \Rightarrow {x^2} + 1 = \frac{1}{6} + 1 \cr & \Rightarrow {x^2} + 1 = \frac{7}{6} \cr} $$

342.

If sin21° = $$\frac{x}{y}{\text{,}}$$  then sec21° - sin69° is equal to?

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

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 $$\eqalign{ & {\text{In }}\vartriangle {\text{ABC sin2}}{{\text{1}}^ \circ }{\text{ = }}\frac{x}{y} \cr & {\text{AB}} = x \cr & {\text{AC}} = y \cr & {\text{BC}} = \sqrt {{y^2} - {x^2}} \cr & \Rightarrow {\text{sec2}}{{\text{1}}^ \circ } - \sin {69^ \circ } \cr & \Rightarrow \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & \Rightarrow \frac{{{{\left( {{\text{AC}}} \right)}^2} - {{\left( {{\text{BC}}} \right)}^2}}}{{\left( {{\text{BC}}} \right)\left( {{\text{AC}}} \right)}} = \frac{{{y^2} - {{\left( {\sqrt {\left( {{y^2} - {x^2}} \right)} } \right)}^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr & \Rightarrow \frac{{{y^2} - {y^2} + {x^2}}}{{y\sqrt {{y^2} + {x^2}} }} = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$

343.

The value of m[sinθ + 2cos∅ + 3sinθ + 4cos∅ + . . . . . . . . + 18cos∅] is a perfect square of an integer, θ = 30°, ∅ = 45° and 150 â‰Â¤ m â‰Â¤ 180. Find the value of m.222

• (A) 161
• (B) 176
• (C) 152
• (D) 168

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 $$\eqalign{ & m\left[ {\sin \theta + 2{{\cos }^2}\phi + 3\sin \theta + 4{{\cos }^2}\phi + \,.....\,18{{\cos }^2}\phi } \right] \cr & \theta = {30^ \circ },\,\,\phi = {45^ \circ } \cr & = m\left[ {\left( {\sin \theta + 3\sin \theta + 5\sin \theta + \,.....\,17\sin \theta } \right) + \left( {2{{\cos }^2}\phi + 4{{\cos }^2}\phi + \,.....\,18{{\cos }^2}\phi } \right)} \right] \cr & \Rightarrow {\text{Sum of odd number}} = {\left( 9 \right)^2}, \cr & {\text{Sum of even number}} = 9 \times 10 \cr & = m\left[ {81\sin \theta + 90{{\cos }^2}\phi } \right] \cr & = m\left[ {81 \times \sin 30 + 90 \times {{\cos }^2}45} \right] \cr & = m\left[ {81 \times \frac{1}{2} + 90 \times \frac{1}{2}} \right] \cr & = m \times \frac{{171}}{2} \cr & {\text{Now, we will check through the option,}} \cr & {\text{Putting }}m = 152 \cr & = 152 \times \frac{{171}}{2} \cr & = 4 \times 19 \times 19 \times 9 \cr & = {\left( {2 \times 19 \times 3} \right)^2} \cr & = {\text{ perfect square}}{\text{. Ans}}{\text{.}} \cr} $$

344.

θ is the positive acute angle and sinθ - cosθ = 0, then the value of secθ + cosecθ is?

• (A) 2
• (B) $$\sqrt 2 $$
• (C) $${\text{2}}\sqrt 2 $$
• (D) $${\text{3}}\sqrt 2 $$

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 $$\eqalign{ & \sin \theta - \cos \theta = 0 \cr & \sin \theta = \cos \theta \cr & \theta = {45^ \circ } \cr & {\text{Then,}} \cr & sec\theta + {\text{cosec}}\theta \cr & = \sqrt 2 + \sqrt 2 \cr & = 2\sqrt 2 \cr} $$

345.

The value of cos²20° + cos²70° is?

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

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 $$\eqalign{ & {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\sin ^2}{70^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = 1 \cr} $$

346.

If sinθ = √3cosθ, 0° θ 90°, then the value of 2sinθ + 6secθ + sinθ secθ + cosecθ is22

• (A) $$\frac{{33 + 10\sqrt 3 }}{6}$$
• (B) $$\frac{{19 + 10\sqrt 3 }}{3}$$
• (C) $$\frac{{19 + 10\sqrt 3 }}{6}$$
• (D) $$\frac{{33 + 10\sqrt 3 }}{3}$$

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 $$\eqalign{ & \sin \theta = \sqrt 3 \cos \theta \cr & \tan \theta = \sqrt 3 \cr & \tan \theta = \tan {60^ \circ } \cr & \theta = {60^ \circ } \cr & 2{\sin ^2}{60^ \circ } + {\sec ^2}{60^ \circ } + \tan {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = 2 \times \frac{3}{4} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{3}{2} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{{11}}{2} + \frac{5}{{\sqrt 3 }} \cr & = \frac{{11\sqrt 3 + 10}}{{2\sqrt 3 }} \cr & = \frac{{33 + 10\sqrt 3 }}{6} \cr} $$

347.

$$\frac{{\left( {2\sin A} \right)\left( {1 + \sin A} \right)}}{{1 + \sin A + \cos A}}$$    is equal to:

• (A) 1 + sinAcosA
• (B) 1 + sinA - cosA
• (C) 1 + cosA - sinA
• (D) 1 - sinAcosA

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 In these type of questions we can go through option- From option B $$\eqalign{ & \frac{{\left( {2\sin A} \right)\left( {1 + \sin A} \right)}}{{\left( {1 + \sin A} \right) + \cos A}} = 1 - \cos A + \sin A \cr & 2\sin A\left( {1 + \sin A} \right) = {\left( {1 + \sin A} \right)^2} - {\cos ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 1 + \sin 2A + 2\sin A - 1 + {\sin ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 2\sin A\left( {1 + \sin A} \right) \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put }}A = {90^ \circ } \cr & \frac{{\left( {2\sin {{90}^ \circ }} \right)\left( {1 + \sin {{90}^ \circ }} \right)}}{{1 + \sin {{90}^ \circ } + \cos {{90}^ \circ }}} = 2 \cr & {\text{Now from option B}} \cr & 1 + \sin A - \cos A \cr & = 1 + \sin {90^ \circ } - \cos {90^ \circ } \cr & = 1 + 1 - 0 \cr & = 2 \cr} $$

348.

What is the value of the expression cos2Acos2B + sin²(A - B) - sin²(A + B)?

• (A) sin(2A - 2B)
• (B) sin(2A + 2B)
• (C) cos(2A + 2B)
• (D) cos(2A - 2B)

Show Answer

 cos2Acos2B + sin2(A - B) - sin2(A + B) = cos2Acos2B + [{sin(A - B) + sin(A + B)}{sin(A - B) - sin(A + B)}] = cos2Acos2B + [(sinAcosB - cosAsinB + sinAcosB + cosAsinB)(sinAcosB - cosAsinB - sinAcosB - cosAsinB)] = cos2Acos2B + [(2sinAcosB) × (-2cosAsinB)] = cos2Acos2B - (2sinAcosA) × (2sinBcosB) = cos2Acos2B - sin2Asin2B = cos(2A + 2B) Alternate: cos2Acos2B + sin2(A - B) - sin2(A + B) = cos2Acos2B + sin2Asin2B = cos(2A + 2B)

349.

If $${\left\{ {\left( {\frac{{\sec \theta - 1}}{{\sec \theta + 1}}} \right)} \right\}^n} = {\text{cosec}}\,\theta - \cot \theta ,$$       then n = ?

• (A) 1
• (B) 0.5
• (C) -1
• (D) -0.5

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 $$\eqalign{ & {\left\{ {\left( {\frac{{\sec \theta - 1}}{{\sec \theta + 1}}} \right)} \right\}^n} = {\text{cosec}}\,\theta - \cot \theta \cr & {\text{Put }}\theta = {45^ \circ } \cr & {\text{So}},\,{\left\{ {\left( {\frac{{\sec {{45}^ \circ } - 1}}{{\sec {{45}^ \circ } + 1}}} \right)} \right\}^n} = {\text{cosec}}\,{45^ \circ } - \cot {45^ \circ } \cr & \Rightarrow {\left\{ {\left( {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \right)} \right\}^n} = \sqrt 2 - 1 \cr & {\text{Rationalize internally on left side, we get}} \cr & \Rightarrow {\left\{ {{{\left( {\sqrt 2 - 1} \right)}^2}} \right\}^n} = \sqrt 2 - 1 \cr & {\text{To equate }}n{\text{ should be }}\frac{1}{2} \cr & {\text{So, }}n = 0.5 \cr} $$

350.

If cosθ = $$\frac{{12}}{{13}},$$ then the value of $$\frac{{\sin \theta \left( {1 - \tan \theta } \right)}}{{\tan \theta \left( {1 + {\text{cosec}}\theta } \right)}}$$    is:

• (A) $$\frac{{25}}{{156}}$$
• (B) $$\frac{{35}}{{234}}$$
• (C) $$\frac{{35}}{{108}}$$
• (D) $$\frac{{25}}{{78}}$$

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 $$\eqalign{ & \cos \theta = \frac{{12}}{{13}} = \frac{B}{H} \cr & P = \sqrt {{{13}^2} - {{12}^2}} = 5\,{\text{cm}} \cr & \frac{{\sin \theta \left( {1 - \tan \theta } \right)}}{{\tan \theta \left( {1 + {\text{cosec}}\theta } \right)}} \cr & = \frac{{\frac{P}{H}\left( {1 - \frac{P}{B}} \right)}}{{\frac{P}{B}\left( {1 + \frac{H}{P}} \right)}} \cr & = \frac{{\frac{5}{{13}}\left( {1 - \frac{5}{{12}}} \right)}}{{\frac{5}{{12}}\left( {1 + \frac{{13}}{5}} \right)}} \cr & = \frac{{\frac{5}{{13}} \times \frac{7}{{12}}}}{{\frac{5}{{12}} \times \frac{{18}}{5}}} \cr & = \frac{{\frac{{35}}{{156}}}}{{\frac{3}{2}}} \cr & = \frac{{35 \times 2}}{{156 \times 3}} \cr & = \frac{{35}}{{234}} \cr} $$