Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{Given,}} \cr & n = \frac{{{\text{cos }}\alpha }}{{{\text{sin }}\beta }}{\text{,}}\,\,\,\,\,\,{\text{m}} = \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} \cr & \Rightarrow {\text{cos }}\alpha = n{\text{ sin }}\beta ,\,and\,\cos \alpha = m\,\cos \beta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{ }}\alpha = {n^2}{\text{ si}}{{\text{n}}^2}{\text{ }}\beta .....(i) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\alpha = {m^2}{\text{ co}}{{\text{s}}^2}{\text{ }}\beta .....(ii) \cr & equation{\text{ }}(i) = (ii) \cr & \Rightarrow {n^2}{\text{ si}}{{\text{n}}^2}\beta = {m^2}{\text{ co}}{{\text{s}}^2}\beta \cr & \Rightarrow {n^2}\left( {{\text{1}} - {\text{ co}}{{\text{s}}^2}\beta } \right) = {m^2}{\text{ co}}{{\text{s}}^2}\beta \cr & \Rightarrow {n^2} - {n^2}{\cos ^2}\beta = {m^2}{\cos ^2}\beta \cr & \Rightarrow {n^2} = {m^2}{\cos ^2}\beta + {n^2}{\cos ^2}\beta \cr & \Rightarrow {n^2} = {\cos ^2}\beta \left( {{m^2} + {n^2}} \right) \cr & \Rightarrow {\cos ^2}\beta = \frac{{{n^2}}}{{{m^2} + {n^2}}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\left( {{\text{tan}}\theta + \cot \theta } \right)^2} = {5^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\cot ^2}\theta + 2{\text{tan}}\theta \cot \theta = 25 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = {\text{25}} - {\text{2}} \cr & \left[ {\because {\text{tan}}\theta .\cot \theta = 1} \right] \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 23 \cr} $$

$$\eqalign{ & {\text{sin A}} = m{\text{ sin B}} \cr & {\text{si}}{{\text{n}}^2}{\text{A}} = {m^2}{\text{si}}{{\text{n}}^2}{\text{B }}......{\text{(i)}} \cr & {\text{Now, ta}}{{\text{n}}^2}{\text{A}} = {n^2}{\text{ta}}{{\text{n}}^2}{\text{B}} \cr & \frac{{{{\sin }^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} = {n^2}\frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & {\text{from equation (i)}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{\frac{{\left( {1 - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)}}{{{m^2}}}}}{{1 - \frac{{{{\sin }^2}{\text{A}}}}{{{m^2}}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr & \Rightarrow {m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}} = {{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}} \cr & \Rightarrow {m^2} - 1 = {\text{co}}{{\text{s}}^2}\theta \left( {{n^2} - 1} \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{A}} = \frac{{{m^2} - 1}}{{{n^2} - 1}} \cr} $$

$$\eqalign{ & {\text{tan }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.tan 4}}{{\text{7}}^ \circ }{\text{.tan 8}}{{\text{6}}^ \circ } \cr & {\text{Here, }} \cr & {\text{tan 8}}{{\text{6}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {4^ \circ }} \right) = {\text{cot }}{{\text{4}}^ \circ } \cr & {\text{tan 4}}{{\text{7}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {{43}^ \circ }} \right) = {\text{cot 4}}{{\text{3}}^ \circ } \cr & {\text{tan }}{{\text{4}}^ \circ }{\text{.cot }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.cot 4}}{{\text{3}}^ \circ } = 1 \cr} $$

tan 1°.tan 2°.tan 3° .......... tan 89°
If A + B = 90°
(tan 1°.tan 89°)(tan 2°.tan 88°) .......... (tan 44°.tan 46°).tan 45°
1 × 1 × 1 .......... 1 × 1 = 1

cot10° .cot20° .cot60°.cot70° .cot80°
[ In cot A.cot B if A + B = 90° then cot A.cot B = 1 ]
(cot10° .cot80°) × (cot20°.cot70°) × cot60°
∴ 1 × 1 × $$\frac{1}{{\sqrt 3 }}$$
= $$\frac{1}{{\sqrt 3 }}$$

$$\cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \frac{1}{{\tan {{72}^ \circ }.{{\sec }^2}{{68}^ \circ }}}} \right)$$
$$ \Rightarrow \cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \cot {{72}^ \circ }.{{\cos }^2}{{68}^ \circ }} \right) $$
$$ \Rightarrow \cot {18^ \circ }.\cot {72^ \circ }\left( {{{\cos }^2}{{22}^ \circ } + {{\cos }^2}{{68}^ \circ }} \right) $$
We know that, cotA.cotB = 1 and cos²A + cos²B = 1( when A + B = 90° )
= 1 × 1 = 1

$$\eqalign{ & {\text{sin}}\left( {2x - {{20}^ \circ }} \right) = {\text{cos}}\left( {2y + {{20}^ \circ }} \right) \cr & \Rightarrow \left( {2x - {{20}^ \circ }} \right) + \left( {2y + {{20}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B, then A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2\left( {x + y} \right) = {90^ \circ } \cr & \Rightarrow x + y = {45^ \circ } \cr & \therefore \tan \left( {x + y} \right) \cr & = \tan {45^ \circ } \cr & = 1 \cr} $$

$$\eqalign{ & {\sin ^2}{5^ \circ } + {\sin ^2}{6^ \circ } + \,.....\,{\sin ^2}{84^ \circ } + {\sin ^2}{85^ \circ } \cr & {\text{Number of terms}} \cr & = \frac{{85 - 5}}{1} + 1 \cr & = 80 + 1 \cr & = 81 \cr} $$
  $$ = \left( {{{\sin }^2}{5^ \circ } + {{\sin }^2}{{85}^ \circ }} \right)\, + $$     $$\,\left( {{{\sin }^2}{6^ \circ } + {{\sin }^2}{{84}^ \circ }} \right)\, + $$     . . . . . upto 40 pairs $$ + $$ middle term
$$\eqalign{ & = {\text{40}} + {\sin ^2}{45^ \circ } \cr & = 40 + \frac{1}{2} \cr & = 40\frac{1}{2} \cr} $$

Alternate:
In case,
when series is in the form of sin²θ or cos²θ,
then sum of series will always be half of number of terms.
$$ = \frac{{81}}{2} = 40\frac{1}{2}$$

Given, (sin²5° + sin²10° + sin²15° + . . . . . . + sin²85° ) + sin²90°
We know that sinθ = cos(90° - θ )
Therefore sin²85° = cos²(90° - 85°) = cos²5°
Similary sin²60° = cos²(90° - 60°) = cos²40°
And we also know that sin²θ + cos²θ = 1
There are 8 pair in given equation
sin²5° + cos²5° = 1
sin²10° + cos²10° = 1
. . . . . . . . . . . .
. . . . . . . . . . . .
sin²40° + cos²40° = 1
i.e. 8 + sin²45° + sin²90°
⇒ 8 + $$\frac{1}{2}$$ + 1 = $$9\frac{1}{2}$$

Short Trick:
 $$\left( {si{n^2}{5^ \circ }\, + \,{{\sin }^2}{{10}^ \circ }\, + \,{{\sin }^2}{{15}^ \circ }\, + \,.....\,{{\sin }^2}{{85}^ \circ }} \right)\, + $$         $$\,{\sin ^2}90$$
$$\eqalign{ & {\text{Number of terms}} \cr & = \left\{ {\left( {\frac{{85 - 5}}{5}} \right) + 1} \right\}{\text{ + 1}} \cr & {\text{ = }}\frac{{17}}{2}{\text{ + 1}} \cr & {\text{Sum of series}} = 9\frac{1}{2} \cr} $$