Practice MCQ Questions and Answer on Trigonometry

351.

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

352.

Solve the following to find its value in terms of trigonometric ratios.
(sinA + cosA)(1 - sinAcosA)

• (A) sin3A + cos3A
• (B) sin2A - cos2A
• (C) [cosA - sinA][sin2A + cos2A]
• (D) sin3A - cos3A

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 (sinA + cosA)(1 - sinA.cosA) = (sinA + cosA)[sin2A + cos2A - sinA.cosA] = (sin3A + cos3A)

353.

What is the value of 5sin²60° + 7sin²45°+ 8cos²45°?

• (A) 25
• (B) $$\frac{57}{4}$$
• (C) $$\frac{45}{4}$$
• (D) 10

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 $$\eqalign{ & 5{\sin ^2}{60^ \circ } + 7{\sin ^2}{45^ \circ } + 8{\cos ^2}{45^ \circ } \cr & = 5{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + 7{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 8{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{{15}}{4} + \frac{7}{2} + \frac{8}{2} \cr & = \frac{{15 + 14 + 16}}{4} \cr & = \frac{{45}}{4} \cr} $$

354.

If θ is a positive acute angle and 4cos²ÃƒÂƒÃ‚ŽÃ‚¸ - 1 = 0, then the value of tan(θ - 15°) is equal to?

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

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 $$\eqalign{ & {\text{4co}}{{\text{s}}^2}\theta - 1 = 0 \cr & \Rightarrow {\text{4co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & \Rightarrow \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \theta = {60^ \circ } \cr & \Rightarrow \tan \left( {\theta - {{15}^ \circ }} \right) \cr & \Rightarrow \tan \left( {{{60}^ \circ } - {{15}^ \circ }} \right) \cr & \Rightarrow \tan {45^ \circ } \cr & \Rightarrow 1 \cr} $$

355.

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

• (A) 60°
• (B) 0°
• (C) 30°
• (D) 90°

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 $$\eqalign{ & \cos \left( {A - B} \right) = \frac{{\sqrt 3 }}{2} \cr & \cos \left( {A - B} \right) = \cos {30^ \circ } \cr & A - B = {30^ \circ }........\left( {\text{i}} \right) \cr & \sec A = 2 \cr & \cos A = \frac{1}{2} = \cos {60^ \circ } \cr & A = {60^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & A = {60^ \circ }\,\& \,B = {30^ \circ } \cr} $$

356.

If θ is a acute angle and sin(θ + 18°) = $$\frac{1}{2}\text{,}$$ then the value of θ in circular measure is?

• (A) $$\frac{\pi }{12}$$ Radians
• (B) $$\frac{\pi }{15}$$ Radians
• (C) $$\frac{2\pi }{5}$$ Radians
• (D) $$\frac{3\pi }{13}$$ Radians

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 $$\eqalign{ & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right){\text{ = }}\frac{1}{2} \cr & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = {\text{sin }}{30^ \circ } \cr & \theta + {18^ \circ } = {30^ \circ } \cr & \therefore \theta = {12^ \circ } \cr & {\text{We know that,}} \cr & {180^ \circ } = \pi \cr & {12^ \circ } = \frac{\pi }{{{{180}^ \circ }}} \times 12 = \frac{\pi }{{15}} \cr} $$

357.

If $$\frac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\frac{1}{7},\theta ,$$     lies in first quadrant, then the value of $$\frac{\text{cosec}\theta +{\cot }^2\theta }{\text{cosec}\theta -{\cot }^2\theta }$$   is:

• (A) $$\frac{19}{5}$$
• (B) $$\frac{37}{19}$$
• (C) $$\frac{22}{3}$$
• (D) $$\frac{37}{12}$$

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 $$\eqalign{ & \frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7} \cr & 7\sec \theta - 7\tan \theta = \sec \theta + \tan \theta \cr & 6\sec \theta = 8\tan \theta \cr & \sin \theta = \frac{6}{8} = \frac{3}{4} = \frac{P}{H} \cr & B = \sqrt {{4^2} - {3^2}} = \sqrt 7 \cr & \Rightarrow \frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }} \cr & = \frac{{\frac{H}{P} + {{\left( {\frac{B}{P}} \right)}^2}}}{{\frac{H}{P} - {{\left( {\frac{B}{P}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}}{{\frac{4}{3} - {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + \frac{7}{9}}}{{\frac{4}{3} - \frac{7}{9}}} \cr & = \frac{{\frac{{12 + 7}}{9}}}{{\frac{{12 - 7}}{9}}} \cr & = \frac{{19}}{5} \cr} $$

358.

If $$\frac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\frac{1}{7},\theta ,$$     lies in first quadrant, then the value of $$\frac{\text{cosec}\theta +{\cot }^2\theta }{\text{cosec}\theta -{\cot }^2\theta }$$   is:

• (A) $$\frac{19}{5}$$
• (B) $$\frac{37}{19}$$
• (C) $$\frac{22}{3}$$
• (D) $$\frac{37}{12}$$

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 $$\eqalign{ & \frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7} \cr & 7\sec \theta - 7\tan \theta = \sec \theta + \tan \theta \cr & 6\sec \theta = 8\tan \theta \cr & \sin \theta = \frac{6}{8} = \frac{3}{4} = \frac{P}{H} \cr & B = \sqrt {{4^2} - {3^2}} = \sqrt 7 \cr & \Rightarrow \frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }} \cr & = \frac{{\frac{H}{P} + {{\left( {\frac{B}{P}} \right)}^2}}}{{\frac{H}{P} - {{\left( {\frac{B}{P}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}}{{\frac{4}{3} - {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + \frac{7}{9}}}{{\frac{4}{3} - \frac{7}{9}}} \cr & = \frac{{\frac{{12 + 7}}{9}}}{{\frac{{12 - 7}}{9}}} \cr & = \frac{{19}}{5} \cr} $$

359.

If secθ + tanθ = p, (p > 1) then $$\frac{\text{cosec}\theta +1}{\text{cosec}\theta -1}=?$$

• (A) 2p2
• (B) $$\frac{p+1}{p-1}$$
• (C) p2
• (D) $$\frac{p-1}{p+1}$$

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 $$\eqalign{ & \sec \theta + \tan \theta = p \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{p}{{\frac{1}{p}}} \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{{{p^2}}}{1} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{{\sec \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{\sec \theta .\cos \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{{\text{cosec }}\theta }}{1} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & {\text{Apply again componendo and dividendo}} \cr & \frac{{{\text{cosec }}\theta + 1}}{{{\text{cosec }}\theta - 1}} = \frac{{{p^2}}}{1} \cr} $$

360.

If sec²A + tan²A = 3, then what is the value of cotA?

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

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Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board