Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{ta}}{{\text{n}}^2}\theta = 1 - {e^2} \cr & \therefore sec\theta + {\text{ta}}{{\text{n}}^3}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .{\text{tan}}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .\frac{{\sin \theta }}{{{\text{cos}}\theta }}.\frac{1}{{\sin \theta }} \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .sec\theta \cr & \Rightarrow sec\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow \sqrt {1 + {\text{ta}}{{\text{n}}^2}\theta } .\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow {\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {1 + 1 - {e^2}} \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {2 - {e^2}} \right)^{\frac{3}{2}}} \cr} $$

$$\eqalign{ & \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a{\text{ }} \cr & \Rightarrow \cos {\text{ }}\alpha = a{\text{ }}\cos {\text{ }}\beta \cr & {\text{On squaring both sides}} \cr & {\cos ^2}\alpha = {a^2}{\cos ^2}\beta \cr & \Rightarrow 1 - {\sin ^2}\alpha = {a^2}\left( {1 - {{\sin }^2}\beta } \right)....(i) \cr & {\text{Again, }}\sin \alpha = {\text{ }}b\sin \beta \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\sin ^2}\alpha = {\text{ }}{b^2}{\sin ^2}\beta \cr & {\text{Put the value of }}{\sin ^2}\alpha {\text{ in equation (i)}} \cr & \Rightarrow {\text{1}} - {b^2}{\sin ^2}\beta = {a^2} - {a^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = {a^2}si{n^2}\beta - {b^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = si{n^2}\beta \left( {{a^2} - {b^2}} \right) \cr & \Rightarrow si{n^2}\beta = \frac{{{a^2} - 1}}{{{a^2} - {b^2}}} \cr} $$

$$\eqalign{ & \sin \theta = 0.7 \cr & \Rightarrow {\sin ^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\left( {0.7} \right)^2} + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow 0.49 + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = 1 - 0.49 \cr & \Rightarrow {\text{cos}}\theta = \sqrt {0.51} \cr} $$

$$\eqalign{ & {\text{2}}\sin \theta + {\text{cos}}\theta = \frac{7}{3} \cr & \Rightarrow \left( {{\text{ta}}{{\text{n}}^2}\theta - {\text{se}}{{\text{c}}^2}\theta } \right) \cr & \Rightarrow \left( {{{\sec }^2}\theta - 1 - {\text{se}}{{\text{c}}^2}\theta } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\left[ {\because 1 + {\text{ta}}{{\text{n}}^2}\theta = {{\sec }^2}\theta } \right] \cr & \Rightarrow - 1 \cr} $$

$$\eqalign{ & {\sec ^2}\theta - \frac{{{{\sin }^2}\theta - 2{{\sin }^4}\theta }}{{{\text{2co}}{{\text{s}}^4}\theta - {\text{co}}{{\text{s}}^2}\theta }} \cr & \Rightarrow {\sec ^2}\theta - \frac{{{{\sin }^2}\theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{{\text{co}}{{\text{s}}^2}\theta \left( {{\text{2co}}{{\text{s}}^2}\theta - 1} \right)}} \cr & \left[ {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta = 2{\text{co}}{{\text{s}}^2}\theta - 1 = 1 - 2{{\sin }^2}\theta } \right] \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

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

Alternate shortcut method :
$$\eqalign{ & {\text{Put }}\theta = {30^ \circ } \cr & \sqrt {\frac{{1 + {\text{sin }}30^ \circ }}{{1 - {\text{sin }}30^ \circ }}} {\text{ + }}\sqrt {\frac{{1 - {\text{sin }}30^ \circ }}{{1 + {\text{sin }}30^ \circ }}} \cr & \Rightarrow \sqrt {\frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}}} + \sqrt {\frac{{1 - \frac{1}{2}}}{{1 + \frac{1}{2}}}} \cr & \Rightarrow \sqrt {\frac{3}{1}} + \sqrt {\frac{1}{3}} \cr & \Rightarrow \frac{4}{{\sqrt 3 }} \cr} $$
Now check with option by puting θ = 30°
$$\eqalign{ & {\text{ = 2 sec 3}}{0^ \circ } \cr & = \frac{{2 \times 2}}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & {\left( {r\cos \theta - \sqrt 3 } \right)^2} + {\left( {r\sin \theta - 1} \right)^2} = 0 \cr & \Rightarrow {\left( {r\cos \theta - \sqrt 3 } \right)^2} = 0,{\text{ }}{\left( {r\sin \theta - 1} \right)^2} = 0 \cr & \Rightarrow r{\text{ cos}}\theta = \sqrt 3 ........(i) \cr & \Rightarrow r\sin \theta = 1.........(ii) \cr & {\text{Squaring and adding equation (i) and (ii)}} \cr & \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 3 + 1 \cr & \Rightarrow {r^2}\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {r^2} = 4 \cr & \Rightarrow r = 2 \cr & {\text{tan}}\theta = \frac{{r\sin \theta }}{{r{\text{ cos}}\theta }} = \frac{1}{{\sqrt 3 }}{\text{ and }}r{\text{ cos}}\theta = \sqrt 3 \cr & \cos \theta = \frac{{\sqrt 3 }}{r}, \cr & \sec \theta = \frac{r}{{\sqrt 3 }} \cr & \frac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan \theta }} \cr & = \frac{{\frac{r}{{\sqrt 3 }} + \frac{r}{{\sqrt 3 }}}}{{\frac{{{r^2}}}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }}}} \cr & = \frac{{r\left( {\frac{2}{{\sqrt 3 }}} \right)}}{{\frac{{{r^2} + 1}}{{\sqrt 3 }}}} \cr & = \frac{{2r}}{{{r^2} + 1}} \cr & = \frac{{2 \times 2}}{{{2^2} + 1}} \cr & = \frac{4}{5} \cr & \cr & {\bf{Alternate:}} \cr & r = 2 \cr & {\text{tan}}\theta = \frac{{r\sin \theta }}{{r{\text{ cos}}\theta }} = \frac{1}{{\sqrt 3 }} \cr & \theta = {30^ \circ } \cr & = \frac{{2\tan {{30}^ \circ } + \sec {{30}^ \circ }}}{{2\sec {{30}^ \circ } + \tan {{30}^ \circ }}} \cr & = \frac{{2 \times \frac{1}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }}}}{{2 \times \frac{2}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }}}} \cr & = \frac{4}{5} \cr} $$

∠A + ∠C = ∠B + ∠D = 180°
∴ ∠A = 180° - ∠C
cosA = cos(180° - C) ⇒ -cosC
Similarly,
cosB = -cosD
⇒ cosA + cosB + cosC + cosD
⇒ cosA + cosB - cosA - cosB = 0

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Alternate Solution :
Put, A = B = C = D = 90°
= cosA + cosB + cosC + cosD
= cos90° + cos90° + cos90° + cos90°
= 0 + 0 + 0 + 0
= 0

$$\eqalign{ & {\text{Given, }}\frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = {\text{2}}\frac{{51}}{{79}} \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = \frac{{209}}{{79}} \cr & {\text{By componendo dividendo}} \cr & \left[ {\frac{a}{b} = \frac{c}{d},{\text{ }}\frac{{a + b}}{{a - b}} = \frac{{c + d}}{{c - d}}} \right] \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta + {\text{sec}}\theta - {\text{tan}}\theta }}{{{\text{sec}}\theta + {\text{tan}}\theta - {\text{sec}}\theta + {\text{tan}}\theta }} = \frac{{209 + 79}}{{209 - 79}} \cr & \Rightarrow \frac{{2sec\theta }}{{2{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{sec\theta }}{{{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{\frac{1}{{{\text{cos}}\theta }}}}{{\frac{{\sin \theta }}{{{\text{cos}}\theta }}}} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{1}{{\sin \theta }} = \frac{{288}}{{130}} \cr & \Rightarrow {\text{Therefore, }}\sin \theta = \frac{{130}}{{288}} \cr & \Rightarrow \sin \theta = \frac{{65}}{{144}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & {\text{1}} + {\text{co}}{{\text{s}}^2}\theta = {\text{3}}\sin \theta .{\text{cos}}\theta {\text{ }}\left( {0 < \theta < \frac{\pi }{2}} \right) \cr & {\text{1}} + {\text{co}}{{\text{s}}^2}\theta = {\text{3}}\sin \theta .{\text{cos}}\theta \cr & {\text{Dividing by }}{\sin ^2}\theta {\text{ in both sides}} \cr & \Rightarrow \frac{{1 + {\text{co}}{{\text{s}}^2}\theta }}{{{{\sin }^2}\theta }} = \frac{{3\sin \theta .{\text{cos}}\theta }}{{{{\sin }^2}\theta }} \cr & \Rightarrow {\text{cose}}{{\text{c}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 3{\text{cot}}\theta \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 3\cot \theta \cr & \left[ {\because 1 + {\text{co}}{{\text{t}}^2}\theta = {\text{cose}}{{\text{c}}^2}\theta } \right] \cr & \Rightarrow 1 + 2{\text{co}}{{\text{t}}^2}\theta = 3\cot \theta \cr & \Rightarrow 2{\text{co}}{{\text{t}}^2}\theta = 3\cot \theta - 1 \cr & {\text{Let }}\theta = {45^ \circ } \cr & \because {\text{cot4}}{{\text{5}}^ \circ } = 1 \cr & \Rightarrow 2{\text{co}}{{\text{t}}^2}{45^ \circ } - 3{\text{cot}}{45^ \circ } + 1 = 0 \cr & \Rightarrow 2 - 3 + 1 = 0 \cr & \Rightarrow 0 = 0 \cr & {\text{Therefore cot}}\theta = {\text{cot}}{45^ \circ } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \cr} $$