Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \frac{{{{\left[ {1 - \tan \left( {{{90}^ \circ } - \theta } \right)} \right]}^2}}}{{\left[ {{{\cos }^2}\left( {{{90}^ \circ } - \theta } \right)} \right]}} - 1 \cr & {\text{By putting }}\theta = {45^ \circ } \cr & \Rightarrow \frac{{{{\left[ {1 - \tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {{{\cos }^2}\left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}} - 1 \cr & \Rightarrow \frac{{{{\left[ {1 - \tan {{45}^ \circ }} \right]}^2}}}{{{{\cos }^2}{{45}^ \circ }}} - 1 \cr & \Rightarrow 0 - 1 \cr & \Rightarrow - 1 \cr & {\text{By satisfying in option A}} \cr & \Rightarrow - \sin 2\theta \cr & \Rightarrow - \sin {90^ \circ } \cr & \Rightarrow - 1 \cr} $$

\[\begin{array}{l} \cos {15^ \circ } - \cos {165^ \circ }\\ = \cos {15^ \circ } - \cos \left( {{{180}^ \circ } - {{15}^ \circ }} \right)\\ = \cos {15^ \circ } + \cos {15^ \circ }\\ = 2\cos {15^ \circ }\\ = 2\left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)\\ = \frac{{\sqrt 3 + 1}}{{\sqrt 2 }}\\ {\bf{Note:}}\\ \left[ \begin{array}{l} \cos {15^ \circ } = \cos \left( {{{45}^ \circ } - {{30}^ \circ }} \right)\\ = \cos {45^ \circ }\cos {30^ \circ } + \sin {45^ \circ }\sin {30^ \circ }\\ = \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} \times \frac{1}{2}\\ = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \end{array} \right] \end{array}\]

$$\eqalign{ & 4 - 2{\sin ^2}\theta - 5\cos \theta = 0 \cr & {\text{Let }}\theta = {60^ \circ } \cr & 4 - 2{\sin ^2}{60^ \circ } - 5\cos {60^ \circ } = 0 \cr & 4 - 2 \times \frac{3}{4} - 5 \times \frac{1}{2} = 0 \cr & 4 - 4 = 0 \cr & \sin \theta + \tan \theta = \sin {60^ \circ } + \tan {60^ \circ } \cr & = \frac{{\sqrt 3 }}{2} + \sqrt 3 \cr & = \frac{{3\sqrt 3 }}{2} \cr} $$

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

$$\eqalign{ & \frac{{{{\cot }^2}A\left( {\sec A - 1} \right)}}{{\left( {1 + \sin A} \right)}} \cr & = \frac{{{{\cos }^2}A\left( {1 - \cos A} \right)}}{{{{\sin }^2}A\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\left( {1 - {{\sin }^2}A} \right)}}{{\left( {1 - {{\cos }^2}A} \right)}} \times \frac{{\left( {1 - \cos A} \right)}}{{\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right).\left( {1 - \cos A} \right)}}{{\left( {1 + \cos A} \right)\left( {1 - \cos A} \right).\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\frac{{\sec A\left( {1 - \sin A} \right)}}{{\left( {\sec A + 1} \right)}}}}{{\sec A}} \cr & = \frac{{{{\sec }^2}A\left( {1 - \sin A} \right)}}{{\left( {1 + \sec A} \right)}} \cr} $$

$$\eqalign{ & \frac{{{{\sin }^2}\theta - 3\sin \theta + 2}}{{{{\cos }^2}\theta }} = 1 \cr & {\text{Put }}\theta = {30^ \circ } \cr & \frac{{{{\sin }^2}{{30}^ \circ } - 3\sin {{30}^ \circ } + 2}}{{{{\cos }^2}{{30}^ \circ }}} = 1 \cr & \frac{1}{4} - \frac{3}{2} + 2 = \frac{3}{4} \cr & \frac{3}{4} = \frac{3}{4} \cr & \cos 2\theta + \sin 3\theta + {\text{cosec}}\,2\theta \cr & = \cos {60^ \circ } + \sin {90^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = 1 + \frac{1}{2} + \frac{2}{{\sqrt 3 }} \cr & = \frac{{9 + 4\sqrt 3 }}{6} \cr} $$

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

$$\eqalign{ & \sec \theta + \tan \theta = 7........\left( {\text{i}} \right) \cr & \sec \theta - \tan \theta = \frac{1}{7}........\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 2\sec \theta = 7 + \frac{1}{7} = \frac{{50}}{7} \cr & \sec \theta = \frac{{25 \to h}}{{7 \to b}}\,\,\,\,\,p = 24 \cr & \sin \theta = \frac{{24}}{{25}} \cr & 5\sin \theta = 5 \times \frac{{24}}{{25}} = \frac{{24}}{5} \cr} $$

$$\eqalign{ & {\sec ^2}\theta + {\tan ^2}\theta = 3\frac{1}{2} \cr & \Rightarrow 1 + {\tan ^2}\theta + {\tan ^2}\theta = \frac{7}{2} \cr & \Rightarrow 2{\tan ^2}\theta = \frac{7}{2} - 1 \cr & \Rightarrow 2{\tan ^2}\theta = \frac{5}{2} \cr & \Rightarrow 3{\tan ^2}\theta = \frac{5}{4} \cr & \Rightarrow \tan \theta = \frac{{\sqrt 5 \to P}}{{2 \to B}} \cr & H = \sqrt {5 + 4} = 3 \cr & \therefore \,\cos \theta + \sin \theta \cr & = \frac{2}{3} + \frac{{\sqrt 5 }}{3} \cr & = \frac{{2 + \sqrt 5 }}{3} \cr} $$