Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{2cos}}\theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & {\text{When,}} \cr & ax \mp by = m \cr & {\text{then, }}bx \mp ay = \sqrt {{a^2} + {b^2} - {m^2}} \cr & 2\cos \theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \cos \theta + 2\sin \theta = \sqrt {4 + 1 - \frac{1}{2}} \cr & \Rightarrow \cos \theta + 2\sin \theta = \frac{3}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & {\sec ^2}\theta + {\text{ta}}{{\text{n}}^2}\theta = 7 \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta = 7 \cr & \Rightarrow 2{\text{ta}}{{\text{n}}^2}\theta = 6 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = 3 \cr & \Rightarrow {\text{tan}}\theta = \sqrt 3 \cr & \Rightarrow \theta = {60^ \circ } \cr & \cr & {\bf{Alternate:}} \cr & {\text{Take help from option }} \cr & {\text{put }}\theta {\text{ = }}{60^ \circ } \cr & {\text{se}}{{\text{c}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{60^ \circ } = 7 \cr & {\left( 2 \right)^2}{\text{ + }}{\left( {\sqrt 3 } \right)^2} = 7 \cr & 7 = 7{\text{ }}\left( {{\text{matched }}} \right) \cr & {\text{So, }}\theta = {60^ \circ } \cr} $$

(sec x sec y + tan x tan y)² - (sec x tan y + tan x sec y)²
= sec²x. sec²y + tan²x. tan²y + 2sec x. sec y. tan x. tan y - sec²x. tan²y - tan²x. sec²y + 2sec x. tan y. tan x sec y
= sec²x [sec²y - tan²y] - tan²x [sec²y - tan²y]
= (sec²x - tan²x) (sec²y - tan²y)
= 1 × 1
= 1

$$\eqalign{ & {\text{According to the question, }} \cr & {\text{A}} = {\sin ^2}\theta + {\text{co}}{{\text{s}}^4}\theta \cr & {\text{Put }}\theta = {90^ \circ }{\text{ for maximum value of A}} \cr & {\text{A}} = {\sin ^2}{90^ \circ } + {\text{co}}{{\text{s}}^4}{90^ \circ } \cr & {\text{A}} = 1 + 0 \cr & {\text{A}} = 1 \cr & {\text{Put }}\theta = {45^ \circ }{\text{ for minimum value of A}} \cr & {\text{A}} = {\sin ^2}{45^ \circ } + {\text{co}}{{\text{s}}^4}{45^ \circ } \cr & {\text{A}} = \frac{1}{2} + \frac{1}{4} \cr & {\text{A}} = \frac{3}{4} \cr & \therefore {\text{A lies in }}\frac{3}{4} \leqslant {\text{A}} \leqslant {\text{1}} \cr} $$

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

$$\eqalign{ & \frac{{{\pi ^c}}}{4}{\text{ = }}\frac{{{{180}^ \circ }}}{4}{\text{ = }}{45^ \circ } \cr & \angle {\text{BAC}} = {180^ \circ } - {75^ \circ } - {45^ \circ } = {60^ \circ } \cr & {180^ \circ } \to \pi \cr & {1^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \cr & {60^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \times {60^ \circ } = \frac{\pi }{3}{\text{ radian}} \cr} $$

$$\cos \theta = \frac{{15 \to {\text{Base}}}}{{17 \to {\text{Hypo}}}}$$

$$\eqalign{ & {\text{Perpendicular = 8}} \cr & \Rightarrow {\text{cot}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{tan}}\theta = \frac{8}{{15}}\left[ {\therefore \tan \theta = \frac{{\text{P}}}{{\text{B}}}} \right] \cr} $$

$$\eqalign{ & XZ - YZ = 2 \cr & h - P = 2 \,........{\text{(i)}} \cr & {h^2} = {(2\sqrt 6 )^2} + {P^2} \cr & {h^2} - {P^2} = {\left( {2\sqrt 6 } \right)^2} \cr & (h - P)(h + P) = 4 \times 6 \cr & (2)(h + P) = 24 \cr & h + P = 12 \,..........{\text{(ii)}} \cr & {\text{Adding eq }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 2h = 14 \cr & h = 7 \,{\text{and }}P = 5 \cr & \therefore secX + tanX \cr & = \frac{h}{{XY}} + \frac{P}{{XY}} \cr & = \frac{7}{{2\sqrt 6 }} + \frac{5}{{2\sqrt 6 }} \cr & = \frac{{12}}{{2\sqrt 6 }} \cr & = \frac{6}{{\sqrt 6 }} \cr & = \sqrt 6 \cr} $$

$$\eqalign{ & {\text{AC}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 \cr & {\text{sin A}} + \operatorname{cotC} \cr & \frac{{{\text{BC}}}}{{{\text{AC}}}} + \frac{{{\text{BC}}}}{{{\text{AB}}}} \cr & \frac{1}{{\sqrt 5 }} + \frac{1}{2} \cr & \Rightarrow \frac{{2 + \sqrt 5 }}{{2\sqrt 5 }} \cr} $$

$$\eqalign{ & {\text{In }}\vartriangle {\text{ABC sin2}}{{\text{1}}^ \circ }{\text{ = }}\frac{x}{y} \cr & {\text{AB}} = x \cr & {\text{AC}} = y \cr & {\text{BC}} = \sqrt {{y^2} - {x^2}} \cr & \Rightarrow {\text{sec2}}{{\text{1}}^ \circ } - \sin {69^ \circ } \cr & \Rightarrow \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & \Rightarrow \frac{{{{\left( {{\text{AC}}} \right)}^2} - {{\left( {{\text{BC}}} \right)}^2}}}{{\left( {{\text{BC}}} \right)\left( {{\text{AC}}} \right)}} = \frac{{{y^2} - {{\left( {\sqrt {\left( {{y^2} - {x^2}} \right)} } \right)}^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr & \Rightarrow \frac{{{y^2} - {y^2} + {x^2}}}{{y\sqrt {{y^2} + {x^2}} }} = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$