Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$${\sin ^2}{25^ \circ } + {\sin ^2}{65^ \circ } + {\text{cose}}{{\text{c}}^2}{57^ \circ } - {\text{ta}}{{\text{n}}^2}{33^ \circ }$$
$$ \Rightarrow \left( {{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }} \right) + $$     $$\left( {{\text{se}}{{\text{c}}^2}{{33}^ \circ } - {\text{ta}}{{\text{n}}^2}{{33}^ \circ }} \right)$$
$$\eqalign{ & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr & {\bf{Note:}} \cr & {\sin ^2}{65^ \circ } = {\sin ^2}\left( {{{90}^ \circ } - {{25}^ \circ }} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{co}}{{\text{s}}^2}{25^ \circ } \cr & {\text{cose}}{{\text{c}}^2}{57^ \circ } = {\operatorname{cosec} ^2}\left( {{{90}^ \circ } - {{33}^ \circ }} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{se}}{{\text{c}}^2}{33^ \circ } \cr} $$

6sin⁴θ + 3cos⁴θ = 2
⇒ 6sin⁴θ + 3(1 - sin²θ )² = 2
⇒ 6sin⁴θ + 3 + 3sin⁴θ - 6sin²θ = 2
⇒ 9sin⁴θ - 6sin²θ + 1 = 0
⇒ (3sin²θ -1)² = 0
⇒ 3sin²θ = 1
⇒ sinθ = $$\frac{1}{{\sqrt 3 }} $$


Now, $${\left[ {7{{\operatorname{cosec} }^6}\theta + 8{{\sec }^6}\theta } \right]^{\frac{1}{3}}}$$
$$\eqalign{ & = {\left[ {7 \times {{(\sqrt 3 )}^6} + 8{{\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)}^6}} \right]^{\frac{1}{3}}} \cr & = {\left[ {7 \times 27 + 8 \times \frac{{27}}{8}} \right]^{\frac{1}{3}}} \cr & = {\left( {8 \times 27} \right)^{\frac{1}{3}}} \cr & = 6 \cr} $$

$$\eqalign{ & = \frac{{2\sin \theta }}{{\cos \theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)}} \cr & = \frac{{2{\text{tan}}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \cr & = \sin2\theta \cr} $$

$$\eqalign{ & \tan \theta = \frac{3}{4} \cr & {\text{cos2}}\theta = \frac{{1 - {\text{ta}}{{\text{n}}^2}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} \cr & {\text{cos2}}\theta = \frac{7}{{25}} \cr} $$

$$\eqalign{ & = \frac{{{\text{sin 6}}{5^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = \frac{{{\text{sin }}\left( {{{90}^ \circ } - {\text{6}}{5^ \circ }} \right)}}{{\cos {{25}^ \circ }}} \cr & \left[ {\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta } \right] \cr & = \frac{{\cos {{25}^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = 1 \cr} $$

Let x = 8cos10°. cos20°. cos40°
Multiply on both side by sin10° and applying formula (2sinθ. cosθ = sin²θ)
⇒ x sin10° = 4 × 2sin10° cos10°. cos20°. cos40°
⇒ x sin10° = 2 × 2sin20°.cos20°. cos40°
⇒ x sin10° = 2 × sin40°. cos40°
⇒ x sin10° = sin80°
⇒ x sin10° = sin(90° - 10°)
⇒ x sin10° = cos10°
then, x = $$\frac{{\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }}}$$
x = cot10°

$$\eqalign{ & = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr & = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr & = 0 \cr} $$

$$\eqalign{ & {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\text{se}}{{\text{c}}^2}{60^ \circ } - {\text{co}}{{\text{t}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{30^ \circ } \cr & = {\left( {\frac{2}{{\sqrt 3 }}} \right)^2} + {\left( 2 \right)^2} - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = \frac{4}{3} + 4 - \frac{1}{3} + \frac{1}{3} \cr & = \frac{{16}}{3} \cr & = 5\frac{1}{3} \cr} $$

$$\eqalign{ & {\text{4}}\angle {\text{A}} = {\text{3}}\angle {\text{B}} = {\text{12}}\angle {\text{C}} \cr & {\text{A}}:{\text{B}}:{\text{C}} = \frac{1}{4}:\frac{1}{3}:\frac{1}{{12}} \cr & {\text{A}}:{\text{B}}:{\text{C}} = 3:4:1 \cr & {\text{Now,}} \cr & 3x + 4x + x = {180^ \circ } \cr & 8x = {180^ \circ } \cr & x = \frac{{{{180}^ \circ }}}{8} \cr & \angle {\text{A}} = 3x = \frac{{{{180}^ \circ }}}{8} \times 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {67.5^ \circ } \cr} $$

$$\eqalign{ & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right){\text{ = }}\frac{1}{2} \cr & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = {\text{sin }}{30^ \circ } \cr & \theta + {18^ \circ } = {30^ \circ } \cr & \therefore \theta = {12^ \circ } \cr & {\text{We know that,}} \cr & {180^ \circ } = \pi \cr & {12^ \circ } = \frac{\pi }{{{{180}^ \circ }}} \times 12 = \frac{\pi }{{15}} \cr} $$