Quantitative Aptitude
TRIGONOMETRY MCQs
Total Questions : 420
| Page 38 of 42 pages
Answer: Option A. -> 1
Shortcut method :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$ $$\left( {1 + {{\cot }^2}\beta } \right)$$ $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{
& \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr
& {\text{Put }} \alpha = \beta = {45^ \circ } \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}{45^ \circ }.{\sin ^2}{45^ \circ }.{\text{cose}}{{\text{c}}^2}{45^ \circ }.{\operatorname{sce} ^2}{45^ \circ } \cr
& \Rightarrow \frac{1}{2}.\frac{1}{2}.2.2 \cr
& \Rightarrow 1 \cr} $$
Alternate :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$ $$\left( {1 + {{\cot }^2}\beta } \right)$$ $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{
& \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \times \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - \beta } \right).{\sin ^2}\alpha .{\text{cose}}{{\text{c}}^2}\beta .{\text{se}}{{\text{c}}^2}\left( {{{90}^ \circ } - \alpha } \right) \cr
& \Rightarrow {\sin ^2}\beta .{\text{cose}}{{\text{c}}^2}\beta .{\sin ^2}\alpha . {\text{cose}}{{\text{c}} ^2}\alpha \cr
& \Rightarrow 1 \cr} $$
Shortcut method :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$ $$\left( {1 + {{\cot }^2}\beta } \right)$$ $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{
& \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr
& {\text{Put }} \alpha = \beta = {45^ \circ } \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}{45^ \circ }.{\sin ^2}{45^ \circ }.{\text{cose}}{{\text{c}}^2}{45^ \circ }.{\operatorname{sce} ^2}{45^ \circ } \cr
& \Rightarrow \frac{1}{2}.\frac{1}{2}.2.2 \cr
& \Rightarrow 1 \cr} $$
Alternate :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$ $$\left( {1 + {{\cot }^2}\beta } \right)$$ $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{
& \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \times \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - \beta } \right).{\sin ^2}\alpha .{\text{cose}}{{\text{c}}^2}\beta .{\text{se}}{{\text{c}}^2}\left( {{{90}^ \circ } - \alpha } \right) \cr
& \Rightarrow {\sin ^2}\beta .{\text{cose}}{{\text{c}}^2}\beta .{\sin ^2}\alpha . {\text{cose}}{{\text{c}} ^2}\alpha \cr
& \Rightarrow 1 \cr} $$
Answer: Option B. -> $$\frac{1}{2}$$
The value of,
cos24° + cos55° + cos125° + cos204° + cos300°
We know that, cos(180° $$ \pm $$ θ) = -cosθ
⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°)
⇒ cos24° + cos55° - cos55° - cos24° + cos60°
⇒ cos60°
⇒ $$\frac{1}{2}$$
The value of,
cos24° + cos55° + cos125° + cos204° + cos300°
We know that, cos(180° $$ \pm $$ θ) = -cosθ
⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°)
⇒ cos24° + cos55° - cos55° - cos24° + cos60°
⇒ cos60°
⇒ $$\frac{1}{2}$$
Answer: Option D. -> $$\frac{{{q^2}}}{{{p^2}}}$$
$$\eqalign{
& {\text{ tan }}{9^ \circ } = \frac{p}{q} \cr
& \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{1 + {{\cot }^2}{{81}^ \circ }}} \cr
& \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{{{\operatorname{cosec} }^2}{{81}^ \circ }}} \cr
& \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}{{81}^ \circ }}} \times {\sin ^2}{81^ \circ } \cr
& \Rightarrow {\text{ta}}{{\text{n}}^2}{81^ \circ } \cr
& \Rightarrow {\text{ta}}{{\text{n}}^2}\left( {{{90}^ \circ } - {9^ \circ }} \right) \cr
& \Rightarrow {\text{co}}{{\text{t}}^2}{9^ \circ } \cr
& \Rightarrow \frac{{{q^2}}}{{{p^2}}} \cr} $$
$$\eqalign{
& {\text{ tan }}{9^ \circ } = \frac{p}{q} \cr
& \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{1 + {{\cot }^2}{{81}^ \circ }}} \cr
& \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{{{\operatorname{cosec} }^2}{{81}^ \circ }}} \cr
& \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}{{81}^ \circ }}} \times {\sin ^2}{81^ \circ } \cr
& \Rightarrow {\text{ta}}{{\text{n}}^2}{81^ \circ } \cr
& \Rightarrow {\text{ta}}{{\text{n}}^2}\left( {{{90}^ \circ } - {9^ \circ }} \right) \cr
& \Rightarrow {\text{co}}{{\text{t}}^2}{9^ \circ } \cr
& \Rightarrow \frac{{{q^2}}}{{{p^2}}} \cr} $$
Answer: Option C. -> $$\frac{\pi }{6}$$
$$\eqalign{
& {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr
& \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr
& \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr
& \Rightarrow 4{\sin ^2}\theta = 1 \cr
& \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr
& \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr
& \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$
$$\eqalign{
& {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr
& \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr
& \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr
& \Rightarrow 4{\sin ^2}\theta = 1 \cr
& \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr
& \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr
& \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$
Answer: Option B. -> 9°
$$\eqalign{
& {\text{tan}}\left( {2\theta + {{45}^ \circ }} \right) = \cot 3\theta \cr
& \left[ {{\text{If tan A}} = {\text{cot B}}} \right] \cr
& ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr
& \Rightarrow \left( {2\theta + {{45}^ \circ }} \right) + 3\theta = {90^ \circ } \cr
& \Rightarrow 5\theta + {45^ \circ } = {90^ \circ } \cr
& \Rightarrow \theta = \frac{{45}}{5} \cr
& \Rightarrow \theta = {9^ \circ } \cr} $$
$$\eqalign{
& {\text{tan}}\left( {2\theta + {{45}^ \circ }} \right) = \cot 3\theta \cr
& \left[ {{\text{If tan A}} = {\text{cot B}}} \right] \cr
& ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr
& \Rightarrow \left( {2\theta + {{45}^ \circ }} \right) + 3\theta = {90^ \circ } \cr
& \Rightarrow 5\theta + {45^ \circ } = {90^ \circ } \cr
& \Rightarrow \theta = \frac{{45}}{5} \cr
& \Rightarrow \theta = {9^ \circ } \cr} $$
Answer: Option C. -> A perfect square of an integer
$${\text{152}}\left( {\sin {{30}^ \circ }\, + \,2{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,3\sin {{30}^ \circ }\, + \,4{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,.....\, + \,17\sin {{30}^ \circ }\, + \,18{\text{co}}{{\text{s}}^2}{{45}^ \circ }} \right)$$
$$ = 152\left\{ {\frac{1}{2}\, + \,2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}\, + \,3\, \times \,\frac{1}{2}\, + \,.....\,\, + \,17 \times \frac{1}{2}\, + \,18{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right\}$$
$$\eqalign{
& = 152\left\{ {\frac{1}{2}\, + \,1\, + \,1\frac{1}{2} + \,.....\,8\frac{1}{2}\, + \,9} \right\} \cr
& = {\text{This is in A}}{\text{.P}}{\text{. where}} \cr
& a = \frac{1}{2},{\text{ }}d = \frac{1}{2},{\text{ }}n = 18 \cr
& = 152\left\{ {\frac{{18}}{2}\left( {2\, \times \,\frac{1}{2}\, + \,\left( {18\, - \,1} \right)\frac{1}{2}} \right)} \right\} \cr
& = 152\left\{ {\frac{{18}}{2}\left( {1\, + \,\frac{{17}}{2}} \right)} \right\} \cr
& = 152 \times 9 \times \frac{{19}}{2} \cr
& = 12996 \cr
& = \sqrt {12996} \cr
& = 114 \cr} $$
$${\text{152}}\left( {\sin {{30}^ \circ }\, + \,2{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,3\sin {{30}^ \circ }\, + \,4{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,.....\, + \,17\sin {{30}^ \circ }\, + \,18{\text{co}}{{\text{s}}^2}{{45}^ \circ }} \right)$$
$$ = 152\left\{ {\frac{1}{2}\, + \,2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}\, + \,3\, \times \,\frac{1}{2}\, + \,.....\,\, + \,17 \times \frac{1}{2}\, + \,18{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right\}$$
$$\eqalign{
& = 152\left\{ {\frac{1}{2}\, + \,1\, + \,1\frac{1}{2} + \,.....\,8\frac{1}{2}\, + \,9} \right\} \cr
& = {\text{This is in A}}{\text{.P}}{\text{. where}} \cr
& a = \frac{1}{2},{\text{ }}d = \frac{1}{2},{\text{ }}n = 18 \cr
& = 152\left\{ {\frac{{18}}{2}\left( {2\, \times \,\frac{1}{2}\, + \,\left( {18\, - \,1} \right)\frac{1}{2}} \right)} \right\} \cr
& = 152\left\{ {\frac{{18}}{2}\left( {1\, + \,\frac{{17}}{2}} \right)} \right\} \cr
& = 152 \times 9 \times \frac{{19}}{2} \cr
& = 12996 \cr
& = \sqrt {12996} \cr
& = 114 \cr} $$
Answer: Option C. -> 2
$$\eqalign{
& {\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right) - 2{\text{ta}}{{\text{n}}^2}\theta \cr
& {\bf{Shortcut method:}} \cr
& {\text{Take, }}\theta = {0^ \circ } \cr
& \Rightarrow {\text{sec }}{0^ \circ }\left( {\frac{{1 + \sin {0^ \circ }}}{{{\text{cos }}{0^ \circ }}} + \frac{{{\text{cos }}{0^ \circ }}}{{1 + \sin {0^ \circ }}}} \right) - 2{\text{ta}}{{\text{n}}^2}{0^ \circ } \cr
& \Rightarrow 1\left( {\frac{{1 + 0}}{1} + \frac{1}{{1 + 0}}} \right) - 0 \cr
& \Rightarrow 2 \cr} $$
$$\eqalign{
& {\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right) - 2{\text{ta}}{{\text{n}}^2}\theta \cr
& {\bf{Shortcut method:}} \cr
& {\text{Take, }}\theta = {0^ \circ } \cr
& \Rightarrow {\text{sec }}{0^ \circ }\left( {\frac{{1 + \sin {0^ \circ }}}{{{\text{cos }}{0^ \circ }}} + \frac{{{\text{cos }}{0^ \circ }}}{{1 + \sin {0^ \circ }}}} \right) - 2{\text{ta}}{{\text{n}}^2}{0^ \circ } \cr
& \Rightarrow 1\left( {\frac{{1 + 0}}{1} + \frac{1}{{1 + 0}}} \right) - 0 \cr
& \Rightarrow 2 \cr} $$
Answer: Option A. -> 43
$$\eqalign{
& x\sin {45^ \circ } = y\operatorname{cosec} {30^ \circ } \cr
& \Rightarrow \frac{x}{y} = \frac{{{\text{cosec 3}}{{\text{0}}^ \circ }}}{{{\text{sin }}{{45}^ \circ }}} \cr
& \Rightarrow \frac{x}{y} = \frac{2}{{\frac{1}{{\sqrt 2 }}}} \cr
& \Rightarrow \frac{x}{y} = \frac{{2\sqrt 2 }}{1} \cr
& \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {\left( {\frac{{2\sqrt 2 }}{1}} \right)^4} \cr
& \Rightarrow \frac{{{x^4}}}{{{y^4}}} = \frac{{64}}{1} \cr
& \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {4^3} \cr} $$
$$\eqalign{
& x\sin {45^ \circ } = y\operatorname{cosec} {30^ \circ } \cr
& \Rightarrow \frac{x}{y} = \frac{{{\text{cosec 3}}{{\text{0}}^ \circ }}}{{{\text{sin }}{{45}^ \circ }}} \cr
& \Rightarrow \frac{x}{y} = \frac{2}{{\frac{1}{{\sqrt 2 }}}} \cr
& \Rightarrow \frac{x}{y} = \frac{{2\sqrt 2 }}{1} \cr
& \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {\left( {\frac{{2\sqrt 2 }}{1}} \right)^4} \cr
& \Rightarrow \frac{{{x^4}}}{{{y^4}}} = \frac{{64}}{1} \cr
& \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {4^3} \cr} $$
Answer: Option A. -> 0
$$\eqalign{
& {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr
& \Rightarrow {\text{se}}{{\text{c}}^2}\alpha - 1 = 1 + 2\left( {{\text{se}}{{\text{c}}^2}\beta - 1} \right) \cr
& \Rightarrow {\sec ^2}\alpha - 1 = 2{\sec ^2}\beta - 1 \cr
& \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}\alpha }} = \frac{2}{{{\text{co}}{{\text{s}}^2}\beta }} \cr
& \Rightarrow \sqrt 2 {\text{cos}}\alpha = {\text{cos}}\beta \cr
& \therefore \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta = 0 \cr
& \cr
& {\bf{Alternate:}} \cr
& {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr
& {\text{Put }}\beta = {45^ \circ } \cr
& {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2.{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr
& {\text{ ta}}{{\text{n}}^2}\alpha = 3 \cr
& {\text{ tan }}\alpha = \sqrt 3 \cr
& \alpha = {60^ \circ } \cr
& {\text{Put }}\alpha = {60^ \circ },{\text{and }}\beta = {45^ \circ } \cr
& = \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta \cr
& = \sqrt 2 {\text{cos }}{60^ \circ } - {\text{cos }}{45^ \circ } \cr
& = \sqrt 2 \times \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr
& = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr
& = 0 \cr} $$
$$\eqalign{
& {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr
& \Rightarrow {\text{se}}{{\text{c}}^2}\alpha - 1 = 1 + 2\left( {{\text{se}}{{\text{c}}^2}\beta - 1} \right) \cr
& \Rightarrow {\sec ^2}\alpha - 1 = 2{\sec ^2}\beta - 1 \cr
& \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}\alpha }} = \frac{2}{{{\text{co}}{{\text{s}}^2}\beta }} \cr
& \Rightarrow \sqrt 2 {\text{cos}}\alpha = {\text{cos}}\beta \cr
& \therefore \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta = 0 \cr
& \cr
& {\bf{Alternate:}} \cr
& {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr
& {\text{Put }}\beta = {45^ \circ } \cr
& {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2.{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr
& {\text{ ta}}{{\text{n}}^2}\alpha = 3 \cr
& {\text{ tan }}\alpha = \sqrt 3 \cr
& \alpha = {60^ \circ } \cr
& {\text{Put }}\alpha = {60^ \circ },{\text{and }}\beta = {45^ \circ } \cr
& = \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta \cr
& = \sqrt 2 {\text{cos }}{60^ \circ } - {\text{cos }}{45^ \circ } \cr
& = \sqrt 2 \times \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr
& = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr
& = 0 \cr} $$
Answer: Option A. -> 2
$$\eqalign{
& {\text{tan}}\theta + {\text{cot}}\theta = 2 \cr
& {\text{Put }}\theta = {45^ \circ } \cr
& 1 + 1 = 2\left( {{\text{matched}}} \right) \cr
& {\text{So, }}\theta = {45^ \circ } \cr
& \Rightarrow {\text{ta}}{{\text{n}}^{100}}{45^ \circ } + {\text{co}}{{\text{t}}^{100}}{45^ \circ } \cr
& \Rightarrow {1^{100}} + {1^{100}} \cr
& \Rightarrow 2 \cr} $$
$$\eqalign{
& {\text{tan}}\theta + {\text{cot}}\theta = 2 \cr
& {\text{Put }}\theta = {45^ \circ } \cr
& 1 + 1 = 2\left( {{\text{matched}}} \right) \cr
& {\text{So, }}\theta = {45^ \circ } \cr
& \Rightarrow {\text{ta}}{{\text{n}}^{100}}{45^ \circ } + {\text{co}}{{\text{t}}^{100}}{45^ \circ } \cr
& \Rightarrow {1^{100}} + {1^{100}} \cr
& \Rightarrow 2 \cr} $$
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