Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

sin²1° + sin²11° + sin²21° + sin²31° + sin²41° + sin²45° + sin²49° + sin²59° + sin²69° + sin²79° + sin²89°
= (sin²1° + sin²89°) + (sin²11° + sin²79°) + (sin²21° + sin²69°) + (sin²31° + sin²59°) + (sin²41° + sin²49°) + sin²45°
= 1 + 1 + 1 + 1 + 1 + $$\frac{1}{2}$$   [sin²A + sin²B = 1. If, A + B = 90°]
= $$5\frac{1}{2}$$

$$\eqalign{ & \cos {20^ \circ } = m{\text{ }} \cr & \cos {70^ \circ } = n \cr & {\text{So,}} \cr & \Leftrightarrow {m^2} + {n^2} = {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & \left[ {{\text{If co}}{{\text{s}}^2}{\text{A + co}}{{\text{s}}^2}{\text{B}} = {\text{1}}} \right] \cr & ({\text{If, A}} + {\text{B}} = {90^ \circ }) \cr & \Leftrightarrow 1 \cr} $$

$$\eqalign{ & {\text{sin}}\left( {{{90}^ \circ } - \theta } \right) + {\text{cos}}\theta = \sqrt 2 {\text{cos}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{cos}}\theta + {\text{cos}}\theta = \sqrt 2 \sin \theta \cr & \Rightarrow \frac{{2\cos \theta }}{{\sin \theta }} = \sqrt 2 \cr & \Rightarrow \cot \theta = \frac{{1 \to {\text{B}}}}{{\sqrt 2 \to {\text{P}}}} \cr & {\text{So, H}} \to \text{alignment} \cr & \therefore {\text{cosec}}\theta = \frac{{\text{H}}}{{\text{P}}} = \frac{{\sqrt 3 }}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{3}{2}} \cr} $$

$$\eqalign{ & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\bf{Shortcut \,\,method:}} \cr & {\text{Put, }}\theta = {60^ \circ } \cr & \Rightarrow {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow {\text{sin }}{60^ \circ } - \cos {60^ \circ } = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} - \frac{1}{2} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 - 1}}{2} = \frac{{\sqrt 3 - 1}}{2}({\text{Matched}}) \cr & Hence, \cr & {\text{sin A}}.cos{\text{A}} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \times \frac{1}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{4} \cr & \cr & {\bf{Alternate:}} \cr & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\text{Squaring both side,}} \cr & \Rightarrow {\text{ si}}{{\text{n}}^2}{\text{ A + }}{\cos ^2}{\text{A}} - {\text{2}}{\text{.sin A}}.cos{\text{A}} = {\left( {\frac{{\sqrt 3 - 1}}{2}} \right)^2} \cr & \Rightarrow 1 - 2{\text{sin A}}.cos{\text{A = }}\frac{{3 + 1 - 2\sqrt 3 }}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = 1 - 2\frac{{\left( {2 - \sqrt 3 } \right)}}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = \frac{{2 - 2 + \sqrt 3 }}{2} \cr & \Rightarrow {\text{sin A}}.cos{\text{A}} = \frac{{\sqrt 3 }}{4} \cr} $$

$$\eqalign{ & \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}{{20}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}{{31}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}\left( {{{90}^ \circ } - {{59}^ \circ }} \right)} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{ta}}{{\text{n}}^2}{{70}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\cot }^2}{{59}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{1}{2} = \frac{2}{m}\left[ {{{\sec }^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1} \right] \cr & (cose{c^2}\theta - {\cot ^2}\theta = 1) \cr & \Rightarrow m = 2 \times 2 \cr & \Rightarrow m = 4 \cr} $$

$$\eqalign{ & \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By componendo and dividendo}} \cr & \Rightarrow \frac{{2{\text{tan}}\theta }}{{2{\text{cos}}\theta }} = \frac{3}{1} \cr & \Rightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} \times \frac{{\sin \theta }}{{{\text{cos}}\theta }} = 3 \cr & \Rightarrow {\sin ^2}\theta = 3{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow {\sin ^2}\theta = 3\left( {1 - {{\sin }^2}\theta } \right) \cr & \Rightarrow 4{\sin ^2}\theta = 3 \cr & \Rightarrow {\sin ^2}\theta \Rightarrow \frac{3}{4} \cr & \Rightarrow {\text{sin }}\theta = \frac{{\sqrt 3 }}{2} \cr & \cr & {\bf{Alternate:}} \cr & \Rightarrow \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By C and D}} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{\cot \theta }} = \frac{3}{1} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = 3 \cr & \Rightarrow {\text{tan}}\theta = \sqrt 3 \cr & \theta = {60^ \circ } \cr & \Rightarrow \sin \theta \cr & \Rightarrow {\text{sin }}{60^ \circ } \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \cr} $$

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
$$\eqalign{ & \frac{a}{b} = \frac{{m + n}}{{m - n}} \cr & \Leftrightarrow \frac{{\sin \theta + \cos \theta }}{{{\text{sin}}\theta - \cos \theta }} = 3 \cr & \Leftrightarrow \sin \theta + {\text{cos}}\theta = 3\sin \theta - 3{\text{cos}}\theta \cr & \Leftrightarrow 2\sin \theta = 4{\text{cos}}\theta \cr & \Leftrightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} = \frac{2}{1} \cr & \Leftrightarrow {\text{tan}}\theta = \frac{2}{1} = \frac{{\text{P}}}{{\text{B}}} \cr & \left[ {\therefore {\text{sin}}\theta = \frac{{\text{P}}}{{\text{H}}} = \frac{2}{{\sqrt 5 }}{\text{ and cos}}\theta = \frac{{\text{B}}}{{\text{H}}} = \frac{1}{{\sqrt 5 }}} \right] \cr} $$

$$\eqalign{ & \Leftrightarrow {\sin ^4}\theta - {\text{co}}{{\text{s}}^4}\theta \cr & \Leftrightarrow \left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right)\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow 1\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^2} - {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} \cr & \Leftrightarrow \frac{4}{5} - \frac{1}{5} \cr & \Leftrightarrow \frac{3}{5} \cr} $$

$${\text{ tan 1}}{5^ \circ }{\text{cot 7}}{5^ \circ } + {\text{tan 7}}{5^ \circ }{\text{cot 1}}{5^ \circ }$$
  $$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$      $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$    $${\text{cot1}}{5^ \circ }$$
$$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$

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

$$\eqalign{ & {\text{ tan}}\left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) = \sqrt 3 \cr & \Rightarrow {\text{tan}}\left( {{{90}^ \circ } - \frac{\theta }{2}} \right) = \sqrt 3 \,\,\left[ {\pi = {{180}^ \circ }} \right] \cr & \Rightarrow \cot \frac{\theta }{2} = \sqrt 3 \cr & \Rightarrow \cot \frac{\theta }{2} = \cot {30^ \circ } \cr & \Rightarrow \frac{\theta }{2} = {30^ \circ } \cr & \Rightarrow \theta = {60^ \circ } \cr & \Rightarrow \cos {60^ \circ } = \frac{1}{2} \cr} $$