Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}{25^ \circ } + {\cos ^2}{35^ \circ } + {\cos ^2}{55^ \circ } \cr & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}\left( {{{90}^ \circ } - {{65}^ \circ }} \right) + \left[ {{{\cos }^2}{{35}^ \circ } + {{\cos }^2}\left( {{{90}^ \circ } - {{35}^ \circ }} \right)} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{65}^ \circ } + {{\cos }^2}{{65}^ \circ }} \right) + \left( {{{\cos }^2}{{35}^ \circ } + si{n^2}{{35}^ \circ }} \right) \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & \sec A - \cos A \cr & = \frac{1}{{\cos A}} - \cos A \cr & = \frac{{1 - {{\cos }^2}A}}{{\cos A}} \cr & = \frac{{{{\sin }^2}A}}{{\cos A}} \cr & = \tan A.\sin A \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta - 3\cos \theta + 2}} = 1 \cr & {\sin ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 1 - {\cos ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 0 = 2{\cos ^2}\theta - 3\cos \theta + 1 \cr & 2{\cos ^2}\theta - 3\cos \theta + 1 = 0 \cr & 2{\cos ^2}\theta - 2\cos \theta - \cos \theta + 1 = 0 \cr & 2\cos \theta \left( {\cos \theta - 1} \right) - 1\left( {\cos \theta - 1} \right) = 0 \cr & \left( {2\cos \theta - 1} \right)\left( {\cos \theta - 1} \right) = 0 \cr & \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \cos \theta = 1 = \cos {0^ \circ } \cr & \Rightarrow \frac{{{{\tan }^2}\frac{\theta }{2} + {{\sin }^2}\frac{\theta }{2}}}{{\tan \theta + \sin \theta }} \cr & = \frac{{{{\tan }^2}\frac{{{{60}^ \circ }}}{2} + {{\sin }^2}\frac{{{{60}^ \circ }}}{2}}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\tan }^2}{{30}^ \circ } + {{\sin }^2}{{30}^ \circ }}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}}}{{\sqrt 3 + \frac{{\sqrt 3 }}{2}}} \cr & = \frac{{\frac{1}{3} + \frac{1}{4}}}{{\frac{{2\sqrt 3 + \sqrt 3 }}{2}}} \cr & = \frac{{\frac{{4 + 3}}{{12}}}}{{\frac{{3\sqrt 3 }}{2}}} \cr & = \frac{{7 \times 2}}{{12 \times 3\sqrt 3 }} \cr & = \frac{7}{{18\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{7\sqrt 3 }}{{54}} \cr} $$

Solution:

 $$\eqalign{ & {\text{7co}}{{\text{s}}^2}\theta + 3{\sin ^2}\theta = 4 \cr & 7{\text{co}}{{\text{s}}^2}\theta + 3\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) - 4 = 0 \cr & 7{\text{co}}{{\text{s}}^2}\theta - 3{\text{co}}{{\text{s}}^2}\theta + 3 - 4 = 0 \cr & 4{\text{co}}{{\text{s}}^2}\theta = 1 \cr & {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & {\text{cos}}\theta = \frac{1}{2} \cr & \cos \theta = {\text{cos}}{60^ \circ } \cr & \theta = {60^ \circ } \cr} $$

Solution:

 $$\eqalign{ & \cos \pi x = {x^2} - x + \frac{5}{4} \cr & = {x^2} - 2 \times x \times \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + \frac{5}{4} \cr & = {\left( {x - \frac{1}{2}} \right)^2} + 1 > 1 \cr & = - 1 \leqslant \cos x \leqslant 1 \cr} $$ ∴ So, value of x is none of the above

Solution:

 $${\text{ }}x{\sin ^2}{60^ \circ } - \frac{3}{2}{\text{sec}}{60^ \circ }{\text{.ta}}{{\text{n}}^2}{30^ \circ } + $$       $$\frac{4}{5}{\sin ^2}{45^ \circ }.$$   $${\text{ta}}{{\text{n}}^2}{60^ \circ }$$   $$ = 0$$ $$ \Rightarrow {\text{ }}x{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - \frac{3}{2} \times {\text{2}} \times {\left( {\frac{1}{{\sqrt 3 }}} \right)^2}{\text{ + }}$$       $$\frac{4}{5}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times $$   $${\left( {\sqrt 3 } \right)^2} = 0$$ $$\eqalign{ & \Rightarrow \frac{{3x}}{4} - \frac{3}{2} \times 2 \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{2} \times 3 = 0 \cr & \Rightarrow \frac{{3x}}{4} - 1 + \frac{6}{5} = 0 \cr & \Rightarrow \frac{{3x}}{4} = 1 - \frac{6}{5} \cr & \Rightarrow \frac{{5 - 6}}{5} \cr & \Rightarrow \frac{{ - 1}}{5} \cr & \therefore x = - \frac{1}{5} \times \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\, = - \frac{4}{{15}} \cr} $$

Solution:

 The value of, cos24° + cos55° + cos125° + cos204° + cos300° We know that, cos(180° $$ \pm $$ θ) = -cosθ ⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°) ⇒ cos24° + cos55° - cos55° - cos24° + cos60° ⇒ cos60° ⇒ $$\frac{1}{2}$$

Solution:

 $$\eqalign{ & {\text{cos}}\theta + \sec \theta = \sqrt 3 \cr & {\text{Cubing both sides}} \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3{\text{cos}}\theta \sec \theta \left( {{\text{cos}}\theta + \sec \theta } \right) = 3\sqrt 3 \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3\sqrt 3 = 3\sqrt 3 \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta = 0 \cr} $$

Solution:

 $${\text{ }}x{\sin ^2}{60^ \circ } - \frac{3}{2}{\text{sec}}{60^ \circ }{\text{.ta}}{{\text{n}}^2}{30^ \circ } + $$       $$\frac{4}{5}{\sin ^2}{45^ \circ }.$$   $${\text{ta}}{{\text{n}}^2}{60^ \circ }$$   $$ = 0$$ $$ \Rightarrow {\text{ }}x{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - \frac{3}{2} \times {\text{2}} \times {\left( {\frac{1}{{\sqrt 3 }}} \right)^2}{\text{ + }}$$       $$\frac{4}{5}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times $$   $${\left( {\sqrt 3 } \right)^2} = 0$$ $$\eqalign{ & \Rightarrow \frac{{3x}}{4} - \frac{3}{2} \times 2 \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{2} \times 3 = 0 \cr & \Rightarrow \frac{{3x}}{4} - 1 + \frac{6}{5} = 0 \cr & \Rightarrow \frac{{3x}}{4} = 1 - \frac{6}{5} \cr & \Rightarrow \frac{{5 - 6}}{5} \cr & \Rightarrow \frac{{ - 1}}{5} \cr & \therefore x = - \frac{1}{5} \times \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\, = - \frac{4}{{15}} \cr} $$