Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & \frac{{\cos {{40}^ \circ } - \cos {{140}^ \circ }}}{{\sin {{80}^ \circ } + \sin {{20}^ \circ }}} \cr & = \frac{{ - 2\sin \left( {\frac{{{{40}^ \circ } + {{140}^ \circ }}}{2}} \right)\sin \left( {\frac{{{{40}^ \circ } - {{140}^ \circ }}}{2}} \right)}}{{2\sin \left( {\frac{{{{80}^ \circ } + {{20}^ \circ }}}{2}} \right)\cos \left( {\frac{{{{80}^ \circ } - {{20}^ \circ }}}{2}} \right)}} \cr & = \frac{{ - 2\sin \left( {\frac{{{{180}^ \circ }}}{2}} \right)\sin \left( { - \frac{{{{100}^ \circ }}}{2}} \right)}}{{2\sin \left( {\frac{{{{100}^ \circ }}}{2}} \right)\cos \left( {\frac{{{{60}^ \circ }}}{2}} \right)}} \cr & = \frac{{ - 2\sin {{90}^ \circ } \times \sin \left( { - {{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{ - 2\sin {{90}^ \circ } \times \left( { - \sin {{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{2\sin {{90}^ \circ } \times \sin {{50}^ \circ }}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{\sin {{90}^ \circ }}}{{\cos {{30}^ \circ }}} \cr & = \frac{1}{{\frac{{\sqrt 3 }}{2}}} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$

Solution:

 $${\text{ tan 1}}{5^ \circ }{\text{cot 7}}{5^ \circ } + {\text{tan 7}}{5^ \circ }{\text{cot 1}}{5^ \circ }$$   $$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$      $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$    $${\text{cot1}}{5^ \circ }$$ $$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$

Solution:

 $$\eqalign{ & \sin \theta = \frac{{{p^2} - 1}}{{{p^2} + 1}} \cr & {\text{Let }}p = 2 \cr} $$ $$\eqalign{ & {\text{Now from option A}} \cr & \frac{{2 \times 2}}{{1 + 4}} = \frac{4}{5} \cr} $$

Solution:

 $${\text{ }}x{\sin ^2}{60^ \circ } - \frac{3}{2}{\text{sec}}{60^ \circ }{\text{.ta}}{{\text{n}}^2}{30^ \circ } + $$       $$\frac{4}{5}{\sin ^2}{45^ \circ }.$$   $${\text{ta}}{{\text{n}}^2}{60^ \circ }$$   $$ = 0$$ $$ \Rightarrow {\text{ }}x{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - \frac{3}{2} \times {\text{2}} \times {\left( {\frac{1}{{\sqrt 3 }}} \right)^2}{\text{ + }}$$       $$\frac{4}{5}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times $$   $${\left( {\sqrt 3 } \right)^2} = 0$$ $$\eqalign{ & \Rightarrow \frac{{3x}}{4} - \frac{3}{2} \times 2 \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{2} \times 3 = 0 \cr & \Rightarrow \frac{{3x}}{4} - 1 + \frac{6}{5} = 0 \cr & \Rightarrow \frac{{3x}}{4} = 1 - \frac{6}{5} \cr & \Rightarrow \frac{{5 - 6}}{5} \cr & \Rightarrow \frac{{ - 1}}{5} \cr & \therefore x = - \frac{1}{5} \times \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\, = - \frac{4}{{15}} \cr} $$

Solution:

 (sinA + cosA)(1 - sinA.cosA) = (sinA + cosA)[sin2A + cos2A - sinA.cosA] = (sin3A + cos3A)

Solution:

 $$\eqalign{ & \cot {53^ \circ } = \frac{x}{y} = \frac{B}{H} \cr & {P^2} = {H^2} - {B^2} \cr & P = \sqrt {{y^2} - {x^2}} \cr & \sec {53^ \circ } + \cot {37^ \circ } \cr & = \sec {53^ \circ } + \tan {53^ \circ } \cr & = \frac{H}{B} + \frac{P}{B} \cr & = \frac{{H + P}}{B} \cr & = \frac{{y + \sqrt {{y^2} - {x^2}} }}{x} \cr} $$

Solution:

 $$\eqalign{ & {\left( {\frac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{1 - \tan \theta }}{{\frac{{\tan \theta - 1}}{{\tan \theta }}}}} \right)^2} + 1 \cr & = \frac{{\left( {1 - \tan \theta } \right){{\tan }^2}\theta }}{{\left( {1 - \tan \theta } \right)}} + 1 \cr & = {\tan ^2}\theta + 1 \cr & = {\sec ^2}\theta \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{1 + \cos \theta - {{\sin }^2}\theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } }}{{\tan \theta + \cot \theta }} \cr & \Rightarrow \frac{{{{\cos }^2}\theta - \cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {\frac{1}{{{{\cos }^2}\theta }} + \frac{1}{{{{\sin }^2}\theta }}} }}{{\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}}} \cr & \Rightarrow \frac{{\cos \theta \left( {\cos \theta + 1} \right)}}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} }}{{\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}}} \cr & \Rightarrow \frac{{\cos \theta }}{{\sin \theta }} \times \frac{{\sqrt {\frac{1}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} }}{{\frac{1}{{\cos \theta \sin \theta }}}} \cr & \Rightarrow \cot \theta \times \frac{{\frac{1}{{\cos \theta \sin \theta }}}}{{\frac{1}{{\cos \theta \sin \theta }}}} \cr & \Rightarrow \cot \theta \cr} $$

Solution:

 $$\eqalign{ & {\text{Put, }}\theta = {60^ \circ } \cr & \Rightarrow \cos \theta > {\cos ^2}\theta \cr & \Rightarrow \cos {60^ \circ } > {\cos ^2}{60^ \circ } \cr & \Rightarrow \frac{1}{2} > \frac{1}{4} \cr & \cos \theta > {\cos ^2}\theta \cr} $$