Practice MCQ Questions and Answer on Trigonometry

Solution:

Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board

Solution:

 $$\eqalign{ & 3{\sec ^2}\theta + \tan \theta - 7 = 0 \cr & \Rightarrow 3\left( {1 + {{\tan }^2}\theta } \right) + \tan \theta - 7 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + \tan \theta - 4 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + 4\tan \theta - 3\tan \theta - 4 = 0 \cr & \Rightarrow \tan \theta \left( {3\tan \theta + 4} \right) - 1\left( {3\tan \theta + 4} \right) = 0 \cr & \Rightarrow \left( {3\tan \theta + 4} \right)\left( {\tan \theta - 1} \right) = 0 \cr & \Rightarrow \tan \theta = 1\,\,\,\,\,\,\,\,\,\therefore \theta = {45^ \circ } \cr & \therefore \,\frac{{2\sin \theta + 3\cos \theta }}{{{\text{cosec}}\,\theta + \sec \theta }} \cr & = \frac{{2\left( {\frac{1}{{\sqrt 2 }}} \right) + 3\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 + \sqrt 2 }} \cr & = \frac{{\frac{5}{{\sqrt 2 }}}}{{2\sqrt 2 }} \cr & = \frac{5}{4} \cr} $$

Solution:

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

Solution:

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

Solution:

 $$\eqalign{ & {\text{A}} + {\text{B}} + {\text{C}} = \pi = {\text{18}}{0^ \circ }{\text{ }} \cr & \Rightarrow \frac{{{\text{A + B}}}}{2} = \frac{{{{180}^ \circ }}}{2} - \frac{{\text{C}}}{2} \cr & \Rightarrow {\text{sin }}\left( {\frac{{{\text{A + B}}}}{2}} \right) \cr & \Rightarrow {\text{sin }}\left( {\frac{\pi }{2} - \frac{{\text{C}}}{2}} \right) \cr & \Rightarrow {\text{cos}}\frac{{\text{C}}}{2} \cr & \cr & {\bf{Similarly:}} \cr & {\text{cos }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = {\text{sin}}\frac{{\text{C}}}{2} \cr & {\text{cot }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \tan \frac{{\text{C}}}{2} \cr & {\text{tan }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \cot \frac{{\text{C}}}{2} \cr & {\text{So, option C is incorrect}}{\text{.}} \cr} $$

Solution:

 $$\eqalign{ & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & {\text{Find }}{\sin ^4}\theta = ? \cr & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & \left( {{\text{by C & D}}} \right) \cr & \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} = \frac{{3 + 1}}{{3 - 1}} \cr & \Rightarrow {\text{tan}}\theta = 2 \cr & {\text{tan}}\theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{1} \cr & \Rightarrow {\sin ^4}\theta \Rightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$

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:

 $$\eqalign{ & {\text{tan}}\theta - \cot \theta = 0 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {45^ \circ } \cr & {\text{tan }}{45^ \circ } - \cot {45^ \circ } = 0 \cr & 1 - 1 = 0 \cr & 0 - 0({\text{matched}}) \cr & So,\theta = {45^ \circ } \cr & \Rightarrow \sin \theta + \cos \theta \cr & \Rightarrow \sin {45^ \circ } + \cos {45^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \sqrt 2 \cr} $$

Solution:

 If in the any question componendo and dividendo already used as $$\frac{{a + b}}{{a - b}} = \frac{m}{n}$$ If second time you also want to apply componendo dividendo rule then result will be $$\eqalign{ & \frac{a}{b} = \frac{{m + n}}{{m - n}} \cr & \Leftrightarrow \frac{{\sin \theta + \cos \theta }}{{{\text{sin}}\theta - \cos \theta }} = 3 \cr & \Leftrightarrow \sin \theta + {\text{cos}}\theta = 3\sin \theta - 3{\text{cos}}\theta \cr & \Leftrightarrow 2\sin \theta = 4{\text{cos}}\theta \cr & \Leftrightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} = \frac{2}{1} \cr & \Leftrightarrow {\text{tan}}\theta = \frac{2}{1} = \frac{{\text{P}}}{{\text{B}}} \cr & \left[ {\therefore {\text{sin}}\theta = \frac{{\text{P}}}{{\text{H}}} = \frac{2}{{\sqrt 5 }}{\text{ and cos}}\theta = \frac{{\text{B}}}{{\text{H}}} = \frac{1}{{\sqrt 5 }}} \right] \cr} $$ $$\eqalign{ & \Leftrightarrow {\sin ^4}\theta - {\text{co}}{{\text{s}}^4}\theta \cr & \Leftrightarrow \left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right)\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow 1\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^2} - {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} \cr & \Leftrightarrow \frac{4}{5} - \frac{1}{5} \cr & \Leftrightarrow \frac{3}{5} \cr} $$

Solution:

 $${\text{ 2}}\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) - {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right){\text{ + 1}}$$ $$ \Rightarrow {\text{2}}\left( {1 - 3{{\sin }^2}\theta {\text{co}}{{\text{s}}^2}\theta } \right) - $$     $${\text{3}}\left( {1 - 2{{\sin }^2}\theta c{\text{o}}{{\text{s}}^2}\theta } \right){\text{ + }}$$     $${\text{1}}$$ $$\eqalign{ & \Rightarrow {\text{ 2}} - 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta - 3 + 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta {\text{ + 1}} \cr & \Rightarrow 2 - 3 + 1 \cr & \Rightarrow 0 \cr} $$