Quantitative Aptitude
TRIGONOMETRY MCQs
Total Questions : 420
| Page 40 of 42 pages
Answer: Option B. -> 0
$$\eqalign{
& = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr
& = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr
& = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr
& = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr
& = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr
& = 0 \cr} $$
$$\eqalign{
& = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr
& = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr
& = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr
& = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr
& = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr
& = 0 \cr} $$
Answer: Option D. -> $${\text{5}}\frac{1}{3}$$
$$\eqalign{
& {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\text{se}}{{\text{c}}^2}{60^ \circ } - {\text{co}}{{\text{t}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{30^ \circ } \cr
& = {\left( {\frac{2}{{\sqrt 3 }}} \right)^2} + {\left( 2 \right)^2} - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr
& = \frac{4}{3} + 4 - \frac{1}{3} + \frac{1}{3} \cr
& = \frac{{16}}{3} \cr
& = 5\frac{1}{3} \cr} $$
$$\eqalign{
& {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\text{se}}{{\text{c}}^2}{60^ \circ } - {\text{co}}{{\text{t}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{30^ \circ } \cr
& = {\left( {\frac{2}{{\sqrt 3 }}} \right)^2} + {\left( 2 \right)^2} - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr
& = \frac{4}{3} + 4 - \frac{1}{3} + \frac{1}{3} \cr
& = \frac{{16}}{3} \cr
& = 5\frac{1}{3} \cr} $$
Answer: Option D. -> cot10°
Let x = 8cos10°. cos20°. cos40°
Multiply on both side by sin10° and applying formula (2sinθ. cosθ = sin2θ)
⇒ x sin10° = 4 × 2sin10° cos10°. cos20°. cos40°
⇒ x sin10° = 2 × 2sin20°.cos20°. cos40°
⇒ x sin10° = 2 × sin40°. cos40°
⇒ x sin10° = sin80°
⇒ x sin10° = sin(90° - 10°)
⇒ x sin10° = cos10°
then, x = $$\frac{{\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }}}$$
x = cot10°
Let x = 8cos10°. cos20°. cos40°
Multiply on both side by sin10° and applying formula (2sinθ. cosθ = sin2θ)
⇒ x sin10° = 4 × 2sin10° cos10°. cos20°. cos40°
⇒ x sin10° = 2 × 2sin20°.cos20°. cos40°
⇒ x sin10° = 2 × sin40°. cos40°
⇒ x sin10° = sin80°
⇒ x sin10° = sin(90° - 10°)
⇒ x sin10° = cos10°
then, x = $$\frac{{\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }}}$$
x = cot10°
Answer: Option C. -> 67.5°
$$\eqalign{
& {\text{4}}\angle {\text{A}} = {\text{3}}\angle {\text{B}} = {\text{12}}\angle {\text{C}} \cr
& {\text{A}}:{\text{B}}:{\text{C}} = \frac{1}{4}:\frac{1}{3}:\frac{1}{{12}} \cr
& {\text{A}}:{\text{B}}:{\text{C}} = 3:4:1 \cr
& {\text{Now,}} \cr
& 3x + 4x + x = {180^ \circ } \cr
& 8x = {180^ \circ } \cr
& x = \frac{{{{180}^ \circ }}}{8} \cr
& \angle {\text{A}} = 3x = \frac{{{{180}^ \circ }}}{8} \times 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {67.5^ \circ } \cr} $$
$$\eqalign{
& {\text{4}}\angle {\text{A}} = {\text{3}}\angle {\text{B}} = {\text{12}}\angle {\text{C}} \cr
& {\text{A}}:{\text{B}}:{\text{C}} = \frac{1}{4}:\frac{1}{3}:\frac{1}{{12}} \cr
& {\text{A}}:{\text{B}}:{\text{C}} = 3:4:1 \cr
& {\text{Now,}} \cr
& 3x + 4x + x = {180^ \circ } \cr
& 8x = {180^ \circ } \cr
& x = \frac{{{{180}^ \circ }}}{8} \cr
& \angle {\text{A}} = 3x = \frac{{{{180}^ \circ }}}{8} \times 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {67.5^ \circ } \cr} $$
Answer: Option B. -> $$\frac{\pi }{{15}}$$ Radians
$$\eqalign{
& {\text{sin}}\left( {\theta + {{18}^ \circ }} \right){\text{ = }}\frac{1}{2} \cr
& {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = {\text{sin }}{30^ \circ } \cr
& \theta + {18^ \circ } = {30^ \circ } \cr
& \therefore \theta = {12^ \circ } \cr
& {\text{We know that,}} \cr
& {180^ \circ } = \pi \cr
& {12^ \circ } = \frac{\pi }{{{{180}^ \circ }}} \times 12 = \frac{\pi }{{15}} \cr} $$
$$\eqalign{
& {\text{sin}}\left( {\theta + {{18}^ \circ }} \right){\text{ = }}\frac{1}{2} \cr
& {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = {\text{sin }}{30^ \circ } \cr
& \theta + {18^ \circ } = {30^ \circ } \cr
& \therefore \theta = {12^ \circ } \cr
& {\text{We know that,}} \cr
& {180^ \circ } = \pi \cr
& {12^ \circ } = \frac{\pi }{{{{180}^ \circ }}} \times 12 = \frac{\pi }{{15}} \cr} $$
Answer: Option C. -> 30°
$$\eqalign{
& {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{50}^ \circ } - x} \right) \cr
& {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{90}^ \circ } - \left( {{{40}^ \circ } + x} \right)} \right) \cr
& {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{sec}}\left( {{{40}^ \circ } + x} \right) \cr
& 4x - {50^ \circ } = {40^ \circ } + x \cr
& 3x = {90^ \circ } \cr
& x = {30^ \circ } \cr} $$
$$\eqalign{
& {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{50}^ \circ } - x} \right) \cr
& {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{90}^ \circ } - \left( {{{40}^ \circ } + x} \right)} \right) \cr
& {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{sec}}\left( {{{40}^ \circ } + x} \right) \cr
& 4x - {50^ \circ } = {40^ \circ } + x \cr
& 3x = {90^ \circ } \cr
& x = {30^ \circ } \cr} $$
Answer: Option C. -> 2
$$\eqalign{
& \pi \sin \theta = 1\,.....(i) \cr
& \pi \cos \theta = 1\,.....(ii) \cr
& {\text{Divide eq}}{\text{. (i) from (ii)}} \cr
& \frac{{\pi \sin \theta }}{{\pi \cos \theta }} = \frac{1}{1} \cr
& \tan \theta = 1 \cr
& \tan \theta = \tan {45^ \circ } \cr
& \theta = {45^ \circ } \cr
& \therefore \sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1 \cr
& = \sqrt 3 \tan \left( {\frac{2}{3} \times {{45}^ \circ }} \right) + 1 \cr
& = \sqrt 3 {\text{ tan}}{30^ \circ } + 1 \cr
& = \sqrt 3 \times \frac{1}{{\sqrt 3 }} + 1 \cr
& = 2 \cr} $$
$$\eqalign{
& \pi \sin \theta = 1\,.....(i) \cr
& \pi \cos \theta = 1\,.....(ii) \cr
& {\text{Divide eq}}{\text{. (i) from (ii)}} \cr
& \frac{{\pi \sin \theta }}{{\pi \cos \theta }} = \frac{1}{1} \cr
& \tan \theta = 1 \cr
& \tan \theta = \tan {45^ \circ } \cr
& \theta = {45^ \circ } \cr
& \therefore \sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1 \cr
& = \sqrt 3 \tan \left( {\frac{2}{3} \times {{45}^ \circ }} \right) + 1 \cr
& = \sqrt 3 {\text{ tan}}{30^ \circ } + 1 \cr
& = \sqrt 3 \times \frac{1}{{\sqrt 3 }} + 1 \cr
& = 2 \cr} $$
Answer: Option B. -> 0
$$\eqalign{
& {\text{In }}\vartriangle {\text{, }}\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C}} = {\text{18}}{0^ \circ }\,....{\text{(i)}} \cr
& {\text{sin A}} = {\text{cos B}} \cr
& {\text{sin A}} = {\text{sin}}\left( {{{90}^ \circ } - {\text{B}}} \right){\text{ }} \cr
& {\text{A}} = {90^ \circ } - {\text{B}} \cr
& {\text{A}} + {\text{B}} = {90^ \circ }\,........(ii) \cr
& {\text{From equation (i) and (ii)}} \cr
& \angle {\text{C}} = {90^ \circ } \cr
& {\text{So, cos C }} = \cos {90^ \circ } = 0 \cr} $$
$$\eqalign{
& {\text{In }}\vartriangle {\text{, }}\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C}} = {\text{18}}{0^ \circ }\,....{\text{(i)}} \cr
& {\text{sin A}} = {\text{cos B}} \cr
& {\text{sin A}} = {\text{sin}}\left( {{{90}^ \circ } - {\text{B}}} \right){\text{ }} \cr
& {\text{A}} = {90^ \circ } - {\text{B}} \cr
& {\text{A}} + {\text{B}} = {90^ \circ }\,........(ii) \cr
& {\text{From equation (i) and (ii)}} \cr
& \angle {\text{C}} = {90^ \circ } \cr
& {\text{So, cos C }} = \cos {90^ \circ } = 0 \cr} $$
Answer: Option A. -> 0
$$\eqalign{
& \sin \theta \times \cos \theta = \frac{1}{2} \cr
& {\text{Multiply by 2 both side}} \cr
& 2\sin \theta \times \cos \theta = 1 \cr
& \sin 2\theta = 1 \cr
& \sin 2\theta = \sin {90^ \circ } \cr
& 2\theta = {90^ \circ } \cr
& \theta = {45^ \circ } \cr
& {\text{So,}}\sin \theta - {\text{cos}}\theta \cr
& = \sin {45^ \circ } - \cos {45^ \circ } \cr
& = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr
& = 0{\text{ }} \cr} $$
$$\eqalign{
& \sin \theta \times \cos \theta = \frac{1}{2} \cr
& {\text{Multiply by 2 both side}} \cr
& 2\sin \theta \times \cos \theta = 1 \cr
& \sin 2\theta = 1 \cr
& \sin 2\theta = \sin {90^ \circ } \cr
& 2\theta = {90^ \circ } \cr
& \theta = {45^ \circ } \cr
& {\text{So,}}\sin \theta - {\text{cos}}\theta \cr
& = \sin {45^ \circ } - \cos {45^ \circ } \cr
& = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr
& = 0{\text{ }} \cr} $$
Answer: Option B. -> 1
$$\eqalign{
& = {\cos ^2}{20^ \circ } + {\cos ^2}{70^ \circ } \cr
& = {\cos ^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\cos ^2}{70^ \circ } \cr
& = 1 \cr} $$
$$\eqalign{
& = {\cos ^2}{20^ \circ } + {\cos ^2}{70^ \circ } \cr
& = {\cos ^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\cos ^2}{70^ \circ } \cr
& = 1 \cr} $$
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