Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & {\text{ta}}{{\text{n}}^2}\theta = 1 - {e^2} \cr & \therefore sec\theta + {\text{ta}}{{\text{n}}^3}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .{\text{tan}}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .\frac{{\sin \theta }}{{{\text{cos}}\theta }}.\frac{1}{{\sin \theta }} \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .sec\theta \cr & \Rightarrow sec\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow \sqrt {1 + {\text{ta}}{{\text{n}}^2}\theta } .\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow {\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {1 + 1 - {e^2}} \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {2 - {e^2}} \right)^{\frac{3}{2}}} \cr} $$

Solution:

 $$\eqalign{ & {\text{The value of }} \cr & {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right) + 2\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) + 12{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta \cr & {\text{Using }}\theta = {0^ \circ } \cr & \because \sin {0^ \circ } = {0^ \circ } \cr & \cos {0^ \circ } = 1 \cr & \Rightarrow 3\left( {0 + {1^4}} \right) + 2\left( {0 + {1^6}} \right) + 12 \times 0 \times 1 \cr & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

Solution:

 $$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$ Divide numerator & denominator by cosθ $$\eqalign{ & = \frac{{\frac{{3\sin \theta }}{{\cos \theta }} + \frac{{2\cos \theta }}{{\cos \theta }}}}{{\frac{{3\sin \theta }}{{\cos \theta }} - \frac{{2\cos \theta }}{{\cos \theta }}}}\left[ {\frac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \cr & = \frac{{3\tan \theta + 2}}{{3\tan \theta - 2}} \cr & {\text{Put value of tan}}\theta \cr & = \frac{{3 \times \frac{4}{3} + 2}}{{3 \times \frac{4}{3} - 2}} \cr & = \frac{6}{2} \cr & = 3 \cr} $$

Solution:

 $$\eqalign{ & \frac{{2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)}}{{{{\cos }^4}\theta - {{\sin }^4}\theta - 2{{\cos }^2}\theta }} \cr & = \frac{{2\left( {1 - 3{{\sin }^2}\theta .{{\cos }^2}\theta } \right) - 3\left( {1 - 2{{\sin }^2}\theta .{{\cos }^2}\theta } \right)}}{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) - 2{{\cos }^2}\theta }} \cr & = \frac{{2 - 6{{\sin }^2}\theta .{{\cos }^2}\theta - 3 + 6{{\sin }^2}\theta .{{\cos }^2}\theta }}{{ - \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}} \cr & = \frac{{ - 1}}{{ - 1}} \cr & = 1 \cr} $$

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:

 $$\eqalign{ & 4{\cos ^2}\theta - 4\sqrt 3 \cos \theta + 3 = 0 \cr & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {30^ \circ }{\text{option A}} \cr & 4{\cos ^2}{30^ \circ } - 4\sqrt 3 \cos {30^ \circ } + 3 = 0 \cr & \Rightarrow 4\left( {\frac{3}{4}} \right) - 4\sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right) + 3 = 0 \cr & \Rightarrow 0 = 0 \cr} $$

Solution:

 $$\eqalign{ & {\text{ tan }}{9^ \circ } = \frac{p}{q} \cr & \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{1 + {{\cot }^2}{{81}^ \circ }}} \cr & \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{{{\operatorname{cosec} }^2}{{81}^ \circ }}} \cr & \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}{{81}^ \circ }}} \times {\sin ^2}{81^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}{81^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\left( {{{90}^ \circ } - {9^ \circ }} \right) \cr & \Rightarrow {\text{co}}{{\text{t}}^2}{9^ \circ } \cr & \Rightarrow \frac{{{q^2}}}{{{p^2}}} \cr} $$

Solution:

 $$\eqalign{ & 1 + {\cot ^2}\theta = \frac{{625}}{{49}} \cr & {\text{cose}}{{\text{c}}^2}\theta = \frac{{625}}{{49}} \cr & {\text{cosec}}\,\theta = \frac{{25}}{7} \cr & \sin \theta = \frac{7}{{25}} \cr & \cos \theta = \frac{{24}}{{25}} \cr & {\left( {\sin \theta + \cos \theta } \right)^{\frac{1}{2}}} \cr & = {\left( {\frac{7}{{25}} + \frac{{24}}{{25}}} \right)^{\frac{1}{2}}} \cr & = {\left( {\frac{{31}}{{25}}} \right)^{\frac{1}{2}}} \cr & = \frac{{\sqrt {31} }}{5} \cr} $$

Solution:

 sin238° + sin252° + sin230° - tan245° = sin238° + cos238° + $${\left( {\frac{1}{2}} \right)^2}$$ - 1   [Here sin2θ + cos2θ = 1] = 1 + $$\frac{1}{4}$$ - 1 = $$\frac{1}{4}$$

Solution:

 $$\eqalign{ & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2cot}}{{61}^ \circ }.\cot {{79}^ \circ }}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2}}\left[ {{\text{cot}}\left( {{{90}^ \circ } - {{29}^ \circ }} \right).\cot \left( {{{90}^ \circ } - {{11}^ \circ }} \right)} \right]}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2tan1}}{{\text{1}}^ \circ }.tan{{29}^ \circ }}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{1}{2} \cr & \Leftrightarrow 2{\text{A}} = {\text{B}} \cr} $$