Practice MCQ Questions and Answer on Trigonometry

Solution:

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

Solution:

 $$\eqalign{ & \frac{{\sin \left( {{{78}^ \circ } + \theta } \right) - \cos \left( {{{12}^ \circ } - \theta } \right) + \left( {{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}{{20}^ \circ }} \right)}}{{\sin {{25}^ \circ }\cos {{65}^ \circ } + \cos {{25}^ \circ }\sin {{65}^ \circ }}} \cr & \frac{{\sin \left( {{{78}^ \circ } + \theta } \right) - \sin \left( {{{78}^ \circ } + \theta } \right) + \left( {{{\tan }^2}{{70}^ \circ } - {{\sec }^2}{{70}^ \circ }} \right)}}{{\sin {{25}^ \circ }\sin {{25}^ \circ } + \cos {{25}^ \circ }\sin {{25}^ \circ }}} \cr & \frac{{ - 1}}{{{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }}} \cr & = - 1 \cr} $$

Solution:

 A + B = C tan(A + B) = tanC $$\frac{{\tan {\text{A}} + \tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}} = \tan {\text{C}}$$ tanA + tanB = tanC - tanAtanBtanC tanAtanBtanC = tanC - tanA - tanB

Solution:

 $$\eqalign{ & {\text{A + B}} = {90^ \circ } \cr & {\text{B}} = {90^ \circ } - {\text{A}} \cr & {\sec ^2}A + se{c^2}{\text{B}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}{\text{B}} \cr} $$   $$ = {\sec ^2}A + se{c^2}\left( {{{90}^ \circ } - {\text{A}}} \right) - $$       $${\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}$$   $$\left( {{{90}^ \circ } - {\text{A}}} \right)$$ $$\eqalign{ & = {\sec ^2}A + {\operatorname{cosec} ^2}{\text{A}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & = \frac{1}{{{{\cos }^2}{\text{A}}}} + \frac{1}{{si{n^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} \times \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}} \cr & = \frac{{{{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} \cr & = \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} \cr & = 0 \cr} $$

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:

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

Solution:

 $$\eqalign{ & b = \frac{c}{{1 + \sin x}} \cr & {\text{Go through option from option B}} \cr & b = \frac{{1 + \sin x - \cos x}}{{1 + \sin x}} \cr & = \frac{{{{\left( {1 + \sin x} \right)}^2} - {{\cos }^2}x}}{{1 + \sin x\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{1 + {{\sin }^2}x + 2{{\sin }^2}x - 1 + {{\sin }^2}x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2{{\sin }^2}x + 2\sin x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2\sin x\left( {\sin x + 1} \right)}}{{1 + \sin x + \cos x}} \cr & = \frac{{2\sin x}}{{1 + \sin x + \cos x}} \cr & = a \cr} $$

Solution:

 $$\eqalign{ & \frac{{\left[ {\sin \left( {90 - A} \right) + \cos \left( {180 - 2A} \right)} \right]}}{{\left[ {\cos \left( {90 - 2A} \right) + \sin \left( {180 - A} \right)} \right]}} \cr & \Rightarrow \frac{{\cos A + \left( { - \cos 2A} \right)}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{\cos A - \cos 2A}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{ - 2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{ - \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{{2A - A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{A}{2}} \right)}}{{\cos \left( {\frac{A}{2}} \right)}} \cr & \Rightarrow \tan \left( {\frac{A}{2}} \right) \cr} $$

Solution:

 $$\eqalign{ & {\text{tan}}\theta + {\text{cot}}\theta = 2 \cr & {\text{Put }}\theta = {45^ \circ } \cr & 1 + 1 = 2\left( {{\text{matched}}} \right) \cr & {\text{So, }}\theta = {45^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^{100}}{45^ \circ } + {\text{co}}{{\text{t}}^{100}}{45^ \circ } \cr & \Rightarrow {1^{100}} + {1^{100}} \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr & \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$