Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {30^ \circ } \cr & \Rightarrow \sec \theta - \tan \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \sec {30^ \circ } - \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{{\sqrt 3 }} - \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}{\text{ }}\left( {{\text{Satisfied}}} \right) \cr & \sec \theta = {30^ \circ } \cr & \Rightarrow \sec \theta .\tan \theta \cr & \Rightarrow \sec {30^ \circ }.\tan {30^ \circ } \cr & \Rightarrow \frac{2}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{3} \cr} $$

(cosec a - sin a) (sec a - cos a) (tan a + cot a)
Put a = 45°
⇒ (cosec45° - sin45°) (sec45° - cos45°) (tan45° + cot45°)
$$\eqalign{ & \Rightarrow \left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)\left( {1 + 1} \right) \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \times 2 \cr & \Rightarrow \frac{1}{2} \times 2 \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {60^ \circ }{\text{option A}} \cr & \Rightarrow 2{\sin ^2}{60^ \circ } = 3\cos {60^ \circ } \cr & \Rightarrow 2{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = 3\left( {\frac{1}{2}} \right) \cr & \Rightarrow \frac{3}{2} = \frac{3}{2}{\text{ }}\left( {{\text{LHS}} = {\text{RHS}}} \right) \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0 \cr & \Rightarrow {a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca = 0 \cr & \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0 \cr & \therefore a = b = c \cr & \vartriangle {\text{ABC}} = {\text{equilateral }}\vartriangle \cr & \therefore \angle {\text{A}} = \angle {\text{B}} = \angle {\text{C}} = {60^ \circ } \cr & {\text{So, }}{\sin ^2}{\text{A}} + {\sin ^2}{\text{B}} + {\sin ^2}{\text{C}} \cr & \Rightarrow {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & \Rightarrow 3{\sin ^2}{60^ \circ } \cr & \Rightarrow 3 \times {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & \Rightarrow \frac{9}{4} \cr} $$

$$\eqalign{ & {\text{tan}}\alpha = 2\left( {{\text{given}}} \right) \cr & \therefore \frac{{{\text{cose}}{{\text{c}}^2}\alpha - {\text{se}}{{\text{c}}^2}\alpha }}{{{\text{cose}}{{\text{c}}^2}\alpha + se{c^2}\alpha }} \cr} $$
(Divide by coses²α both in N and D)
$$\eqalign{ & = \frac{{1 - {\text{ta}}{{\text{n}}^2}\alpha }}{{1 + {\text{ta}}{{\text{n}}^2}\alpha }} \cr & = \frac{{1 - {{\left( 2 \right)}^2}}}{{1 + {{\left( 2 \right)}^2}}} \cr & = - \frac{3}{5} \cr} $$

$$\eqalign{ & \sin \left( {\theta + {{30}^ \circ }} \right) = \frac{3}{{\sqrt {12} }} = \frac{3}{{2\sqrt 3 }} \cr & = \frac{{\sqrt 3 }}{2} = \sin \left( {\theta + {{30}^ \circ }} \right) = \sin {60^ \circ } \cr & \therefore \theta = {30^ \circ } \cr & {\text{co}}{{\text{s}}^2}\theta = {\text{co}}{{\text{s}}^2}{30^ \circ } \cr & = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & = \frac{3}{4} \cr} $$

$$\eqalign{ & 4{\cos ^2}\theta - 4\sqrt 3 \cos \theta + 3 = 0 \cr & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {30^ \circ }{\text{option A}} \cr & 4{\cos ^2}{30^ \circ } - 4\sqrt 3 \cos {30^ \circ } + 3 = 0 \cr & \Rightarrow 4\left( {\frac{3}{4}} \right) - 4\sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right) + 3 = 0 \cr & \Rightarrow 0 = 0 \cr} $$

$$\eqalign{ & {\text{According to the question ,}} \cr & {\text{ se}}{{\text{c}}^4}{\text{A}}\left( {1 - {{\sin }^4}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}} \cr & {\text{Put A}} = {45^ \circ } \cr & \Rightarrow {\text{ se}}{{\text{c}}^4}{45^ \circ }\left( {1 - {{\sin }^4}{{45}^ \circ }} \right) - 2{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr & \Rightarrow 4\left( {1 - \frac{1}{4}} \right) - 2 \cr & \Rightarrow 4 \times \frac{3}{4} - {\text{2}} \cr & \Rightarrow 3 - 2 \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{2}{3} \cr & \left[ {{a^4} - {b^4} = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)} \right] \cr & \Rightarrow \left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = \frac{2}{3} \cr & \left[ {{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right] \cr & \Rightarrow 1 \times \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = \frac{2}{3} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) = \frac{2}{3} \cr & \left[ {{{\sin }^2}\theta = 1 - {\text{co}}{{\text{s}}^2}\theta } \right] \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - 1 = \frac{2}{3} \cr} $$

$$\eqalign{ & \sin \theta - \cos \theta = \frac{7}{{13}} = \alpha \cr & {\text{When, }} \cr & ax + by = m\,......(i) \cr & bx - ay = n\,......(ii) \cr} $$
By adding these two equations after making square on both sides we get,
$$\eqalign{ & \left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right) = {m^2} + {n^2} \cr & {\text{In the same process}} \cr & \sin \theta \pm \cos \theta = a \cr & {\text{Then,}}\sin \theta \pm \cos \theta = \sqrt {2 - {a^2}} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {2 - {{\left( {\frac{7}{{13}}} \right)}^2}} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {2 - \left( {\frac{{49}}{{169}}} \right)} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {\frac{{289}}{{169}}} \cr & \Rightarrow \sin \theta + \cos \theta = \frac{{17}}{{13}} \cr} $$