Practice MCQ Questions and Answer on Trigonometry

If $\sqrt{3}$ tanÃÂÃÂÃÂÃÂ¸ = 3sinÃÂÃÂÃÂÃÂ¸, then the value of (sin²ÃÂÃÂÃÂÃÂ¸ - cos²ÃÂÃÂÃÂÃÂ¸) is?

(A) 1
(B) 3
(C) $\frac{1}{3}$
(D) None of these

Solution:

 $$\eqalign{ & {\text{ }}\sqrt 3 \tan \theta = 3\sin \theta \cr & {\bf{Shortcut method:}} \cr & \Rightarrow {\text{ }}\sqrt 3 \frac{{\sin \theta }}{{\cos \theta }} = 3\sin \theta \cr & \Rightarrow \frac{{\sqrt 3 }}{{\cos \theta }} = 3 \cr & \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{3} \cr & {\text{then perpendicular}} = \sqrt 6 \cr} $$ $$\eqalign{ & \Rightarrow \left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow {\left( {\frac{P}{H}} \right)^2} - {\left( {\frac{B}{H}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\sqrt 6 }}{3}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{3}} \right)^2} \cr & \Rightarrow \frac{6}{9} - \frac{3}{9} \cr & \Rightarrow \frac{1}{3} \cr} $$

In circular measure, the value of the angle 11ÃÂÃÂÃÂÃÂ° 15' is?

(A) $\frac{π^{c}}{16}$
(B) $\frac{π^{c}}{8}$
(C) $\frac{π^{c}}{4}$
(D) $\frac{π^{c}}{12}$

Solution:

 $$\eqalign{ & {\text{1}}{{\text{1}}^ \circ }{\text{15'}} = {\text{1}}{{\text{1}} + }\frac{{{\text{15}}}}{{60}} = 11 + \frac{1}{4} = \frac{{{{45}^ \circ }}}{4} \cr & {\text{We know }}\pi {\text{ radian}} = {180^ \circ } \cr & {1^ \circ } = \left( {\frac{\pi }{{{{180}^ \circ }}}} \right){\text{radian,}} \cr & \frac{{{{45}^ \circ }}}{4} = \frac{\pi }{{{{180}^ \circ }}} \times \frac{{{{45}^ \circ }}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{{\pi ^c}}}{{16}}{\text{ }} \cr} $$

If $A \times tan (θ + 150^{\circ})$ ÃÂÃÂ ÃÂÃÂ = $B \times \tan$ $(θ - 60^{\circ}),$ ÃÂÃÂ the value of $\frac{A - B}{A + B}$ ÃÂÃÂ is?

(A) $- \frac{\sin θ}{2}$
(B) $\frac{\sin 2 θ}{2}$
(C) $\frac{\cos 2 θ}{2}$
(D) 0

Solution:

 $$\eqalign{ & {\text{A}} \times {\text{tan}}\left( {\theta + {{150}^ \circ }} \right) = {\text{B}} \times \tan \left( {\theta - {{60}^ \circ }} \right) \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {\theta - {{60}^ \circ }} \right)}}{{\tan \left( {\theta + {{150}^ \circ }} \right)}} \cr & {\text{Put }}\theta = {90^ \circ } \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {{{90}^ \circ } - {{60}^ \circ }} \right)}}{{\tan \left( {{{90}^ \circ } + {{150}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan \left( {{{180}^ \circ } + {{60}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{1}{3} \cr & {\text{then, }}\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - \frac{4}{2} \cr & \Rightarrow \frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - 2 \cr & \Rightarrow \frac{{{\text{A}} - {\text{B}}}}{{{\text{A}} + {\text{B}}}} = - \frac{1}{2} \cr & {\text{Put in option (i)}} \cr & - \frac{{\sin {{90}^ \circ }}}{2} = - \frac{1}{2} \cr & {\text{So, option (A) is correct }} \cr} $$

If $\sec θ - \tan θ = \frac{1}{\sqrt{3}},$ ÃÂÃÂ ÃÂÃÂ the value of $\sec θ$ . $\tan θ$ ÃÂÃÂ = ?

(A) $\frac{4}{\sqrt{3}}$
(B) $\frac{2}{\sqrt{3}}$
(C) $\frac{2}{3}$
(D) $\frac{1}{\sqrt{3}}$

Solution:

 $$\eqalign{ & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {30^ \circ } \cr & \Rightarrow \sec \theta - \tan \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \sec {30^ \circ } - \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{{\sqrt 3 }} - \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}{\text{ }}\left( {{\text{Satisfied}}} \right) \cr & \sec \theta = {30^ \circ } \cr & \Rightarrow \sec \theta .\tan \theta \cr & \Rightarrow \sec {30^ \circ }.\tan {30^ \circ } \cr & \Rightarrow \frac{2}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{3} \cr} $$

If $s e c θ = x + \frac{1}{4 x}$ ÃÂÃÂ $(0^{\circ} θ 90^{\circ})$ ÃÂÃÂ then $s e c θ$ ÃÂÃÂ + $tan θ$ ÃÂÃÂ is equal to?

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

Solution:

 $$\eqalign{ & sec\theta = x + \frac{1}{{4x}} \cr & {\text{tan}}\theta = \sqrt {{\text{sec}}\theta - 1} \cr & {\text{tan}}\theta = \sqrt {{{\left[ {\frac{{4{x^2} + 1}}{{4x}}} \right]}^2} - 1} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} + 1} \right)}^2} - {{\left( {4x} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 + 8{x^2} - 16{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 - 8{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \frac{{\left( {4{x^2} - 1} \right)}}{{4x}} \cr & \therefore sec\theta + {\text{tan}}\theta \cr & = \frac{{4{x^2} + 1}}{{4x}} + \frac{{4{x^2} - 1}}{{4x}} \cr & = \frac{{4{x^2} + 1 + 4{x^2} - 1}}{{4x}} \cr & = \frac{{8{x^2}}}{{4x}} \cr & = 2x \cr & \cr & {\bf{Alternate:}} \cr & sec\theta = x + \frac{1}{{4x}} \cr & {\text{Put }}x = 1 \cr & sec\theta = 1 + \frac{1}{4} = \frac{5}{4} = \frac{{\text{H}}}{{\text{B}}} \cr & \tan \theta = \frac{{\text{P}}}{{\text{B}}} = \frac{3}{4} \cr & {\text{Now, }} \cr & sec\theta + {\text{tan}}\theta \cr & = \frac{5}{4} + \frac{3}{4} \cr & = \frac{{5 + 3}}{4} \cr & = \frac{8}{4} \cr & = 2 \times 1 \cr & = 2x\left( {x = 1} \right) \cr} $$

The value of $\frac{2 (\sin^{6} θ + \cos^{6} θ) - 3 (\sin^{4} θ + \cos^{4} θ)}{\cos^{4} θ - \sin^{4} θ - 2 \cos^{2} θ} is:$

(A) 1
(B) 2
(C) -2
(D) -1

Solution:

 $$\eqalign{ & \frac{{2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)}}{{{{\cos }^4}\theta - {{\sin }^4}\theta - 2{{\cos }^2}\theta }} \cr & = \frac{{2\left( {1 - 3{{\sin }^2}\theta .{{\cos }^2}\theta } \right) - 3\left( {1 - 2{{\sin }^2}\theta .{{\cos }^2}\theta } \right)}}{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) - 2{{\cos }^2}\theta }} \cr & = \frac{{2 - 6{{\sin }^2}\theta .{{\cos }^2}\theta - 3 + 6{{\sin }^2}\theta .{{\cos }^2}\theta }}{{ - \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}} \cr & = \frac{{ - 1}}{{ - 1}} \cr & = 1 \cr} $$

In sin(A - B) = $\frac{1}{2}$ and cos(A + B) = $\frac{1}{2}$ where A > B > 0 and A + B is an acute angle, then the value of B is?

(A) $\frac{π}{6}$
(B) $\frac{π}{12}$
(C) $\frac{π}{4}$
(D) $\frac{π}{2}$

Solution:

 $$\eqalign{ & {\text{sin}}\left( {{\text{A}} - {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} - {\text{B}} = {\text{3}}{0^ \circ }} \right) \cr & {\text{cos}}\left( {{\text{A}} + {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} + {\text{B}} = {{60}^ \circ }} \right) \cr & {\text{Adding}}\,{\text{both}}\,{\text{side}} \cr & \Rightarrow \left( {{\text{A}} - {\text{B}}} \right) + \left( {{\text{A + B}}} \right) = {30^ \circ } + {60^ \circ } \cr & \Rightarrow 2{\text{A}} = {90^ \circ } \cr & \Rightarrow {\text{A}} = {45^ \circ } \cr & \because {\text{A}} - {\text{B}} = {30^ \circ } \cr & {\text{B}} = {\text{A}} - {30^ \circ } \cr & \Rightarrow {45^ \circ } - {30^ \circ } \cr & \Rightarrow {15^ \circ } \cr & \Leftrightarrow \frac{{15 \times \pi }}{{180}} = \frac{\pi }{{12}}({\text{radian}}) \cr} $$

What is the value of $\frac{[\sin (90 - A) + \cos (180 - 2 A)]}{[\cos (90 - 2 A) + \sin (180 - A)]} ?$

(A) sin $(\frac{A}{2})$ cosA
(B) cot $(\frac{A}{2})$
(C) tan $(\frac{A}{2})$
(D) sinAcos $(\frac{A}{2})$

Solution:

 $$\eqalign{ & \frac{{\left[ {\sin \left( {90 - A} \right) + \cos \left( {180 - 2A} \right)} \right]}}{{\left[ {\cos \left( {90 - 2A} \right) + \sin \left( {180 - A} \right)} \right]}} \cr & \Rightarrow \frac{{\cos A + \left( { - \cos 2A} \right)}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{\cos A - \cos 2A}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{ - 2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{ - \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{{2A - A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{A}{2}} \right)}}{{\cos \left( {\frac{A}{2}} \right)}} \cr & \Rightarrow \tan \left( {\frac{A}{2}} \right) \cr} $$

If A = 10ÃÂÃÂÃÂÃÂ°, what is the value of: $\frac{12 \sin 3 A + 5 \cos (5 A - 5^{\circ})}{9 \sin \frac{9 A}{2} - 4 \cos (5 A + 10^{\circ})} ?$

(A) $\frac{6 \sqrt{2} + 5}{9 - 2 \sqrt{2}}$
(B) $\frac{6 \sqrt{2} - 5}{9 - 2 \sqrt{2}}$
(C) $\frac{9 - 2 \sqrt{2}}{6 \sqrt{2} + 5}$
(D) $\frac{6 \sqrt{2} + 5}{9 + 2 \sqrt{2}}$

Solution:

 $$\eqalign{ & A = {10^ \circ } \cr & \frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin 3 \times {{10}^ \circ } + 5\cos \left( {5 \times {{10}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{9 \times {{10}^ \circ }}}{2} - 4\cos \left( {5 \times {{10}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos \left( {{{50}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{{{90}^ \circ }}}{2} - 4\cos \left( {{{50}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos {{45}^ \circ }}}{{9\sin {{45}^ \circ } - 4\cos {{60}^ \circ }}} \cr & = \frac{{12 \times \frac{1}{2} + 5 \times \frac{1}{{\sqrt 2 }}}}{{9 \times \frac{1}{{\sqrt 2 }} - 4 \times \frac{1}{2}}} \cr & = \frac{{6 + \frac{5}{{\sqrt 2 }}}}{{\frac{9}{{\sqrt 2 }} - 2}} \cr & = \frac{{6\sqrt 2 + 5}}{{9 - 2\sqrt 2 }} \cr} $$

10.

The value of tan²ÃÂÃÂÃÂÃÂ¸ + cot²ÃÂÃÂÃÂÃÂ¸ - sec²ÃÂÃÂÃÂÃÂ¸cosec²ÃÂÃÂÃÂÃÂ¸ is equal to:

(A) -2
(B) 1
(C) 0
(D) -1

Solution:

 tan2θ + cot2θ - sec2θ.cosec2θ Put θ = 45° = tan245° + cot245° - sec245°.cosec245° = 1 + 1 - (√2)2.(√2)2 = 2 - 4 = -2

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Practice MCQ Questions and Answer on Trigonometry

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