Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

As we know triplet 7, 24, 25
AC - BC = 25 - 24 = 1
sinC = $$\frac{7}{{25}}$$

$$\eqalign{ & {\sin ^2}{30^ \circ }.{\cos ^2}{45^ \circ } + 2{\tan ^2}{30^ \circ } - {\sec ^2}{60^ \circ } \cr & = {\left( {\frac{1}{2}} \right)^2}.{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - {\left( 2 \right)^2} \cr & = \frac{1}{4} \times \frac{1}{2} + 2{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - {\left( 2 \right)^2} \cr & = \frac{1}{8} + \frac{2}{3} - 4 \cr & = \frac{{3 + 16 - 96}}{{24}} \cr & = - \frac{{77}}{{24}} \cr} $$

$$\eqalign{ & 3\tan \theta = 2\sqrt 3 \sin \theta \cr & \cos \theta = \frac{{\sqrt 3 }}{2} = \frac{B}{H} \cr & P = \sqrt {{3^2} + {2^2}} = \sqrt {13} \cr & \cos \theta = \cos {30^ \circ } \cr & \theta = {30^ \circ } \cr & \Rightarrow \frac{{{\text{cose}}{{\text{c}}^2}2\theta + {{\cot }^2}2\theta }}{{{{\sin }^2}\theta + {{\tan }^2}2\theta }} \cr & = \frac{{{\text{cose}}{{\text{c}}^2}{{60}^ \circ } + {{\cot }^2}{{60}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\tan }^2}{{60}^ \circ }}} \cr & = \frac{{\frac{4}{3} + \frac{1}{3}}}{{\frac{1}{4} + 3}} \cr & = \frac{{\frac{5}{3}}}{{\frac{{13}}{4}}} \cr & = \frac{{20}}{{39}} \cr} $$

$$\eqalign{ & \Rightarrow \frac{{\sin \phi + \tan \theta }}{{\sin \phi - \tan \theta }} \cr & = \frac{{\frac{5}{{13}} + \frac{{12}}{9}}}{{\frac{5}{{13}} - \frac{{12}}{9}}} \cr & = \frac{{15 + 52}}{{39}} \times \frac{{39}}{{15 - 52}} \cr & = - \frac{{67}}{{37}} \cr} $$

$$\eqalign{ & \cot \theta = \sqrt 7 \cr & \frac{{{\text{cose}}{{\text{c}}^2}\theta - {{\sec }^2}\theta }}{{{\text{cose}}{{\text{c}}^2}\theta + {{\sec }^2}\theta }} \cr & = \frac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta + {{\sin }^2}\theta }} \cr & {\text{Divide by }}{\sin ^2}\theta \cr & = \frac{{\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} - \frac{{{{\sin }^2}\theta }}{{{{\sin }^2}\theta }}}}{{\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} + \frac{{{{\sin }^2}\theta }}{{{{\sin }^2}\theta }}}} \cr & = \frac{{{{\cot }^2}\theta - 1}}{{{{\cot }^2}\theta + 1}} \cr & = \frac{{7 - 1}}{{7 + 1}} \cr & = \frac{6}{8} \cr & = \frac{3}{4} \cr} $$

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

$$\eqalign{ & {\sin ^2}{64^ \circ } + \cos {64^ \circ }\sin {26^ \circ } + 2\cos {43^ \circ }{\text{cosec}}\,{47^ \circ } \cr & \Rightarrow {\sin ^2}{64^ \circ } + \cos {64^ \circ }\sin \left( {{{90}^ \circ } - {{64}^ \circ }} \right) + 2\cos {43^ \circ }{\text{cosec}}\left( {{{90}^ \circ } - {{43}^ \circ }} \right) \cr & \Rightarrow {\sin ^2}{64^ \circ } + \cos {64^ \circ }\cos {64^ \circ } + 2\cos {43^ \circ }\sec {43^ \circ } \cr & \Rightarrow {\sin ^2}{64^ \circ } + {\cos ^2}{64^ \circ } + 2\cos {43^ \circ }\left( {\frac{1}{{\cos {{43}^ \circ }}}} \right) \cr & \Rightarrow 1 + 2 \cr & \Rightarrow 3 \cr} $$