Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

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

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

$$\eqalign{ & \sin \left( {A + B} \right) = \cos \left( {A + B} \right) \cr & \frac{{\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right)}} = 1 \cr & \tan \left( {A + B} \right) = 1 \cr & \tan \left( {A + B} \right) = \tan {45^ \circ } \cr & A + B = {45^ \circ } \cr & A = {45^ \circ } - B \cr & \tan A = \tan {45^ \circ } - \tan B \cr & \tan A = \frac{{\tan {{45}^ \circ } - \tan B}}{{1 + \tan {{45}^ \circ }\tan B}} \cr & \tan A = \frac{{1 - \tan B}}{{1 + \tan B}} \cr} $$

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

$$\eqalign{ & \cos \left( {A - B} \right) = \frac{{\sqrt 3 }}{2} \cr & \cos \left( {A - B} \right) = \cos {30^ \circ } \cr & A - B = {30^ \circ }........\left( {\text{i}} \right) \cr & \sec A = 2 \cr & \cos A = \frac{1}{2} = \cos {60^ \circ } \cr & A = {60^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & A = {60^ \circ }\,\& \,B = {30^ \circ } \cr} $$

$$\eqalign{ & \frac{{2{{\cos }^3}\theta - \cos \theta }}{{\sin \theta - 2{{\sin }^3}\theta }} \cr & = \frac{{\cos \theta \left[ {2{{\cos }^2}\theta - 1} \right]}}{{\sin \theta \left[ {1 - 2{{\sin }^2}\theta } \right]}} \cr & = \frac{{\cos \theta \times \cos 2\theta }}{{\sin \theta \times \cos 2\theta }} \cr & = \cot \theta \cr} $$

$$\eqalign{ & \frac{{\sec \theta \left( {\sin \theta - 2{{\sin }^3}\theta } \right)}}{{2{{\cos }^3}\theta - \cos \theta }} \cr & \frac{{\sec \theta .\sin \theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}} \cr & \frac{{\sec \theta .\sin \theta \times \cos 2\theta }}{{\cos \theta \times \cos 2\theta }} \cr & \sec \theta .\tan \theta \cr} $$

$$\eqalign{ & \frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}} \cr & \Rightarrow \frac{{1 + 2{{\tan }^2}x - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{1 + 2\left( {{{\sec }^2}x - 1} \right) - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{2{{\sec }^2}x - 1 - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{2\sec x\left( {\sec x - \tan x} \right)}}{{\sec x - \tan x}} - \frac{1}{{\sec x - \tan x}} \cr & \Rightarrow 2\sec x - \frac{1}{{\sec x - \tan x}} \cr & \Rightarrow 2\sec x - \frac{1}{{\sec x - \tan x}} \times \frac{{\sec x + \tan x}}{{\sec x + \tan x}} \cr & \Rightarrow 2\sec x - \frac{{\sec x + \tan x}}{{{{\sec }^2}x - {{\tan }^2}x}} \cr & \Rightarrow 2\sec x - \sec x - \tan x \cr & \Rightarrow \sec x - \tan x \cr & \cr & {\bf{Alternative:}} \cr & \frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}} \cr & {\text{By putting }}x = {45^ \circ }{\text{ in equation}} \cr & \Rightarrow \frac{{1 + 2{{\tan }^2}{{45}^ \circ } - 2\sec {{45}^ \circ }\tan {{45}^ \circ }}}{{\sec {{45}^ \circ } - \tan {{45}^ \circ }}} \cr & \Rightarrow \frac{{1 + 2 - 2\sqrt 2 }}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{{3 - 2\sqrt 2 }}{{\sqrt 2 - 1}} \times \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \cr & \Rightarrow 3\sqrt 2 - 2\sqrt 2 - 1 \cr & \Rightarrow \sqrt 2 \cr & {\text{By satisfying in options}} \cr & \Rightarrow \sec x - \tan x \Rightarrow \sqrt 2 - 1 \cr} $$

$$\eqalign{ & \frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ }\tan {{310}^ \circ }}} \cr & = \frac{{\tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right) - \tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}}{{1 + \tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right)\tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}} \cr & = \frac{{ - \tan {{40}^ \circ } + \cot {{40}^ \circ }}}{{1 + \tan {{40}^ \circ } \times \cot {{40}^ \circ }}} \cr & = \frac{{\cot {{40}^ \circ } - \tan {{40}^ \circ }}}{{1 + 1}} \cr & = \frac{{\frac{1}{\alpha } - \alpha }}{2} \cr & = \frac{{1 - {\alpha ^2}}}{{2\alpha }} \cr} $$

$$\eqalign{ & \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & A = {60^ \circ },\,\,x = {15^ \circ }{\text{ given,}} \cr & \Rightarrow \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{60}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}} = \frac{a}{b} \cr & \Rightarrow \frac{{\cos {{75}^ \circ }}}{{\cos {{135}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{\sin {{15}^ \circ }}}{{ - \sin {{45}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{{{{\sin }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = 2{\sin ^2}{15^ \circ }........\left( {\text{i}} \right) \cr & \Rightarrow \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{180}^ \circ }} \right)}} = \frac{b}{c} \cr & \Rightarrow \frac{{ - \sin {{45}^ \circ }}}{{ - \cos {{15}^ \circ }}} = \frac{b}{c} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = \frac{{{{\cos }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = 2{\cos ^2}{15^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{Now, }}{\left( {\frac{a}{b} + \frac{c}{b}} \right)^2} + {\left( {\frac{a}{b} - \frac{c}{b}} \right)^2} \cr & = 2\left[ {\frac{{{a^2}}}{{{b^2}}} + \frac{{{c^2}}}{{{b^2}}}} \right] \cr & = 2\left[ {2{{\sin }^2}{{15}^ \circ } + 2{{\cos }^2}{{15}^ \circ }} \right] \cr & = 4 \cr} $$