Quantitative Aptitude
TRIGONOMETRY MCQs
Total Questions : 420
| Page 36 of 42 pages
Answer: Option B. -> 1
$$\eqalign{
& {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr
& \Rightarrow {\text{co}}{{\text{s}}^4}\theta = 1 - {\cos ^2}\theta \cr
& \Rightarrow {\text{co}}{{\text{s}}^4}\theta = {\sin ^2}\theta \cr
& \Rightarrow {\cos ^2}\theta .{\cos ^2}\theta = {\sin ^2}\theta \cr
& \Rightarrow {\cos ^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\theta = {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^4}\theta \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr} $$
$$\eqalign{
& {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr
& \Rightarrow {\text{co}}{{\text{s}}^4}\theta = 1 - {\cos ^2}\theta \cr
& \Rightarrow {\text{co}}{{\text{s}}^4}\theta = {\sin ^2}\theta \cr
& \Rightarrow {\cos ^2}\theta .{\cos ^2}\theta = {\sin ^2}\theta \cr
& \Rightarrow {\cos ^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\theta = {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^4}\theta \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr} $$
Answer: Option B. -> tan57° cot37°
$$\eqalign{
& \frac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr
& = \frac{{cot{{33}^ \circ } + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr
& = \frac{{\frac{1}{{\tan {{33}^ \circ }}} + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \frac{1}{{\tan {{53}^ \circ }}}}} \cr
& = \frac{{1 + \tan {{53}^ \circ }.\tan {{33}^ \circ }}}{{\tan {{33}^ \circ }.\tan {{53}^ \circ } + 1}} \times \frac{{\tan {{53}^ \circ }}}{{\tan {{33}^ \circ }}} \cr
& = \tan {53^ \circ }.cot{33^ \circ } \cr
& = \cot {37^ \circ }.\tan {57^ \circ } \cr} $$
$$\eqalign{
& \frac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr
& = \frac{{cot{{33}^ \circ } + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr
& = \frac{{\frac{1}{{\tan {{33}^ \circ }}} + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \frac{1}{{\tan {{53}^ \circ }}}}} \cr
& = \frac{{1 + \tan {{53}^ \circ }.\tan {{33}^ \circ }}}{{\tan {{33}^ \circ }.\tan {{53}^ \circ } + 1}} \times \frac{{\tan {{53}^ \circ }}}{{\tan {{33}^ \circ }}} \cr
& = \tan {53^ \circ }.cot{33^ \circ } \cr
& = \cot {37^ \circ }.\tan {57^ \circ } \cr} $$
Answer: Option C. -> 1
(secA - cosA)2 + (cosecA - sinA)2 - (cotA - tanA)2
= (sec2A + cos2A - 2secA.cosA) + (coses2A + sin2A - 2cosecA.sinA) - (cot2A + tan2A - 2cotA.tanA)
= sec2A - tan2A + cos2A + sin2A + coses2A - cot2A - 2
= 3 - 2
= 1
Alternate shortcut method:
(secA - cosA)2 + (cosecA - sinA)2 - (cotA - tanA)2
Put θ = 45°
= (sec45° - cos45°)2 + (cosec45° - sin45°)2 - (cot45° - tan45°)2
$$\eqalign{
& = {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {1 - 1} \right)^2} \cr
& = \frac{1}{2} + \frac{1}{2} - 0 \cr
& = 1 \cr} $$
(secA - cosA)2 + (cosecA - sinA)2 - (cotA - tanA)2
= (sec2A + cos2A - 2secA.cosA) + (coses2A + sin2A - 2cosecA.sinA) - (cot2A + tan2A - 2cotA.tanA)
= sec2A - tan2A + cos2A + sin2A + coses2A - cot2A - 2
= 3 - 2
= 1
Alternate shortcut method:
(secA - cosA)2 + (cosecA - sinA)2 - (cotA - tanA)2
Put θ = 45°
= (sec45° - cos45°)2 + (cosec45° - sin45°)2 - (cot45° - tan45°)2
$$\eqalign{
& = {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {1 - 1} \right)^2} \cr
& = \frac{1}{2} + \frac{1}{2} - 0 \cr
& = 1 \cr} $$
Answer: Option C. -> 3.0
$$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$
Divide numerator & denominator by cosθ
$$\eqalign{
& = \frac{{\frac{{3\sin \theta }}{{\cos \theta }} + \frac{{2\cos \theta }}{{\cos \theta }}}}{{\frac{{3\sin \theta }}{{\cos \theta }} - \frac{{2\cos \theta }}{{\cos \theta }}}}\left[ {\frac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \cr
& = \frac{{3\tan \theta + 2}}{{3\tan \theta - 2}} \cr
& {\text{Put value of tan}}\theta \cr
& = \frac{{3 \times \frac{4}{3} + 2}}{{3 \times \frac{4}{3} - 2}} \cr
& = \frac{6}{2} \cr
& = 3 \cr} $$
$$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$
Divide numerator & denominator by cosθ
$$\eqalign{
& = \frac{{\frac{{3\sin \theta }}{{\cos \theta }} + \frac{{2\cos \theta }}{{\cos \theta }}}}{{\frac{{3\sin \theta }}{{\cos \theta }} - \frac{{2\cos \theta }}{{\cos \theta }}}}\left[ {\frac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \cr
& = \frac{{3\tan \theta + 2}}{{3\tan \theta - 2}} \cr
& {\text{Put value of tan}}\theta \cr
& = \frac{{3 \times \frac{4}{3} + 2}}{{3 \times \frac{4}{3} - 2}} \cr
& = \frac{6}{2} \cr
& = 3 \cr} $$
Answer: Option C. -> 29
(x + 5)° + (2x - 3)° + (3x + 4)° = 180°
(Sum of all angles in triangle is 180°)
⇒ 6x + 6° = 180°
⇒ (x + 1) = 30°
⇒ x = 29°
(x + 5)° + (2x - 3)° + (3x + 4)° = 180°
(Sum of all angles in triangle is 180°)
⇒ 6x + 6° = 180°
⇒ (x + 1) = 30°
⇒ x = 29°
Answer: Option B. -> sec223°
According to the question,
$$2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}{67^ \circ } - {\sin ^2}{23^ \circ } - $$ $${\sin ^2}{67^ \circ } - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$ $${\sin ^2}{23^ \circ } - $$ $${\sin ^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ ta}}{{\text{n}}^2}{23^ \circ } - $$ $$\left( {{{\sin }^2}{{23}^ \circ } + co{s^2}{{23}^ \circ }} \right) - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$\eqalign{
& \Rightarrow \frac{2}{{{\text{co}}{{\text{s}}^2}{{23}^ \circ }}} - 1 - {\text{co}}{{\text{t}}^2}{67^ \circ } \cr
& \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) \cr
& \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{ta}}{{\text{n}}^2}{23^ \circ } \cr
& \Rightarrow 2se{c^2}{23^ \circ } - \left( {{\text{1}} + {\text{ta}}{{\text{n}}^2}{{23}^ \circ }} \right) \cr
& \Rightarrow 2se{c^2}{23^ \circ } - {\sec ^2}{23^ \circ } \cr
& \Rightarrow se{c^2}{23^ \circ } \cr} $$
According to the question,
$$2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}{67^ \circ } - {\sin ^2}{23^ \circ } - $$ $${\sin ^2}{67^ \circ } - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$ $${\sin ^2}{23^ \circ } - $$ $${\sin ^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ ta}}{{\text{n}}^2}{23^ \circ } - $$ $$\left( {{{\sin }^2}{{23}^ \circ } + co{s^2}{{23}^ \circ }} \right) - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$\eqalign{
& \Rightarrow \frac{2}{{{\text{co}}{{\text{s}}^2}{{23}^ \circ }}} - 1 - {\text{co}}{{\text{t}}^2}{67^ \circ } \cr
& \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) \cr
& \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{ta}}{{\text{n}}^2}{23^ \circ } \cr
& \Rightarrow 2se{c^2}{23^ \circ } - \left( {{\text{1}} + {\text{ta}}{{\text{n}}^2}{{23}^ \circ }} \right) \cr
& \Rightarrow 2se{c^2}{23^ \circ } - {\sec ^2}{23^ \circ } \cr
& \Rightarrow se{c^2}{23^ \circ } \cr} $$
Answer: Option D. -> tanθ + cotθ + 1
$$\eqalign{
& \frac{{{\text{tan}}\theta }}{{1 - {\text{cot}}\theta }} + \frac{{{\text{cot}}\theta }}{{1 - {\text{tan}}\theta }} \cr
& \Rightarrow \frac{{{\text{tan}}\theta }}{{1 - \frac{1}{{\tan \theta }}}} + \frac{{\frac{1}{{\tan \theta }}}}{{1 - {\text{tan}}\theta }} \cr
& \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta }}{{{\text{tan}}\theta - 1}} + \frac{1}{{{\text{tan}}\theta \left( {1 - \tan \theta } \right)}} \cr
& \Rightarrow \frac{{{\text{tan}}^2\theta }}{{{\text{tan}}\theta - 1}} - \frac{1}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr
& \Rightarrow \frac{{{\text{ta}}{{\text{n}}^3}\theta - 1}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr
& \Rightarrow \frac{{\left( {{\text{tan}}\theta - 1} \right)\left( {{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}} \right)}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr
& \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}}}{{{\text{tan}}\theta }} \cr
& \Rightarrow {\text{tan}}\theta + \cot \theta + {\text{1}} \cr} $$
$$\eqalign{
& \frac{{{\text{tan}}\theta }}{{1 - {\text{cot}}\theta }} + \frac{{{\text{cot}}\theta }}{{1 - {\text{tan}}\theta }} \cr
& \Rightarrow \frac{{{\text{tan}}\theta }}{{1 - \frac{1}{{\tan \theta }}}} + \frac{{\frac{1}{{\tan \theta }}}}{{1 - {\text{tan}}\theta }} \cr
& \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta }}{{{\text{tan}}\theta - 1}} + \frac{1}{{{\text{tan}}\theta \left( {1 - \tan \theta } \right)}} \cr
& \Rightarrow \frac{{{\text{tan}}^2\theta }}{{{\text{tan}}\theta - 1}} - \frac{1}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr
& \Rightarrow \frac{{{\text{ta}}{{\text{n}}^3}\theta - 1}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr
& \Rightarrow \frac{{\left( {{\text{tan}}\theta - 1} \right)\left( {{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}} \right)}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr
& \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}}}{{{\text{tan}}\theta }} \cr
& \Rightarrow {\text{tan}}\theta + \cot \theta + {\text{1}} \cr} $$
Answer: Option A. -> 1
$$\eqalign{
& x = a{\text{ }}\sec \theta .\cos \phi \cr
& y = b{\text{ }}\sec \theta .sin\phi \cr
& z = c{\text{ tan}}\theta \cr
& \frac{x}{a} = \sec \theta .\cos \phi \cr
& \frac{y}{b} = \sec \theta .sin\phi \cr
& \frac{z}{c} = {\text{tan}}\theta \cr
& \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr
& \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow 1 \cr} $$
$$\eqalign{
& x = a{\text{ }}\sec \theta .\cos \phi \cr
& y = b{\text{ }}\sec \theta .sin\phi \cr
& z = c{\text{ tan}}\theta \cr
& \frac{x}{a} = \sec \theta .\cos \phi \cr
& \frac{y}{b} = \sec \theta .sin\phi \cr
& \frac{z}{c} = {\text{tan}}\theta \cr
& \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr
& \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr
& \Rightarrow 1 \cr} $$
Answer: Option D. -> 3
$$\eqalign{
& \frac{1}{{1 + {{\cot }^2}\theta }} + \frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} + 2{\sin ^2}\theta \cr
& \Rightarrow \frac{1}{{{{\operatorname{cosec} }^2}\theta }} + \frac{3}{{{{\sec }^2}\theta }} + 2{\sin ^2}\theta \cr
& \Rightarrow {\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta \cr
& \Rightarrow 3\left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) \cr
& \Rightarrow 3\left( 1 \right) \cr
& \Rightarrow 3 \cr} $$
$$\eqalign{
& \frac{1}{{1 + {{\cot }^2}\theta }} + \frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} + 2{\sin ^2}\theta \cr
& \Rightarrow \frac{1}{{{{\operatorname{cosec} }^2}\theta }} + \frac{3}{{{{\sec }^2}\theta }} + 2{\sin ^2}\theta \cr
& \Rightarrow {\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta \cr
& \Rightarrow 3\left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) \cr
& \Rightarrow 3\left( 1 \right) \cr
& \Rightarrow 3 \cr} $$
Answer: Option B. -> 2
$$\eqalign{
& {\bf{Shortcut \,\, method:}} \cr
& = \left( {1 + \cot \theta - \operatorname{cosec} \theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) \cr
& \left[ {put,\theta = {{45}^ \circ }} \right] \cr} $$
$$ = \left( {1 + \cot {{45}^ \circ } - \operatorname{cosec} {{45}^ \circ }} \right)$$ $$\left( {1 + \tan {{45}^ \circ } + \sec {{45}^ \circ }} \right)$$
$$\eqalign{
& = \left( {1 + 1 - \sqrt 2 } \right)\left( {1 + 1 + \sqrt 2 } \right) \cr
& = \left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right) \cr
& = \left[ {{2^2} - {{\left( {\sqrt 2 } \right)}^2}} \right]\left[ {\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right] \cr
& = 4 - 2 \cr
& = 2 \cr} $$
$$\eqalign{
& {\bf{Shortcut \,\, method:}} \cr
& = \left( {1 + \cot \theta - \operatorname{cosec} \theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) \cr
& \left[ {put,\theta = {{45}^ \circ }} \right] \cr} $$
$$ = \left( {1 + \cot {{45}^ \circ } - \operatorname{cosec} {{45}^ \circ }} \right)$$ $$\left( {1 + \tan {{45}^ \circ } + \sec {{45}^ \circ }} \right)$$
$$\eqalign{
& = \left( {1 + 1 - \sqrt 2 } \right)\left( {1 + 1 + \sqrt 2 } \right) \cr
& = \left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right) \cr
& = \left[ {{2^2} - {{\left( {\sqrt 2 } \right)}^2}} \right]\left[ {\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right] \cr
& = 4 - 2 \cr
& = 2 \cr} $$
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