Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & {\text{sin}}\left( {{{60}^ \circ } - \theta } \right) = {\text{cos}}\left( {\psi - {{30}^ \circ }} \right) \cr & \Rightarrow \left( {{{60}^ \circ } - \theta } \right) + \left( {\psi - {{30}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B}}\,{\text{then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {\psi - \theta } \right) = {90^ \circ } - {30^ \circ } \cr & \Rightarrow \left( {\psi - \theta } \right) = {60^ \circ } \cr & \Rightarrow \tan \left( {\psi - \theta } \right) = \tan {60^ \circ } \cr & \Rightarrow \tan {60^ \circ } = \sqrt 3 \cr} $$

Solution:

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

Solution:

 \[\begin{array}{l} \cos {15^ \circ } - \cos {165^ \circ }\\ = \cos {15^ \circ } - \cos \left( {{{180}^ \circ } - {{15}^ \circ }} \right)\\ = \cos {15^ \circ } + \cos {15^ \circ }\\ = 2\cos {15^ \circ }\\ = 2\left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)\\ = \frac{{\sqrt 3 + 1}}{{\sqrt 2 }}\\ {\bf{Note:}}\\ \left[ \begin{array}{l} \cos {15^ \circ } = \cos \left( {{{45}^ \circ } - {{30}^ \circ }} \right)\\ = \cos {45^ \circ }\cos {30^ \circ } + \sin {45^ \circ }\sin {30^ \circ }\\ = \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} \times \frac{1}{2}\\ = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \end{array} \right] \end{array}\]

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:

 (cos6θ + sin6θ - 1)(tan2θ + cot2θ + 2) + 3 Put θ = 45° = (cos645° + sin645° - 1)(tan245° + cot245° + 2) + 3 $$\eqalign{ & = \left( {\frac{1}{8} + \frac{1}{8} - 1} \right)\left( {1 + 1 + 2} \right) + 3 \cr & = - \frac{3}{4} \times 4 + 3 \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & {\left( {\sec \phi - \tan \phi } \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = {\left( {\frac{1}{{\cos \phi }} - \frac{{\sin \phi }}{{\cos \phi }}} \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = {\left( {\frac{{1 - \sin \phi }}{{\cos \phi }}} \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = \frac{{{{\left( {1 - \sin \phi } \right)}^2}{{\left( {1 + \sin \phi } \right)}^2}}}{{{{\cos }^2}\phi }} \times \frac{1}{{{{\cos }^2}\phi }} \cr & = \frac{{{{\left( {1 - {{\sin }^2}\phi } \right)}^2}}}{{{{\cos }^4}\phi }} \cr & = \frac{{{{\cos }^4}\phi }}{{{{\cos }^4}\phi }} \cr & = 1 \cr} $$

Solution:

 (sinA + cosA)(1 - sinA.cosA) = (sinA + cosA)[sin2A + cos2A - sinA.cosA] = (sin3A + cos3A)

Solution:

 $$\eqalign{ & 3{\sec ^2}\theta + \tan \theta - 7 = 0 \cr & \Rightarrow 3\left( {1 + {{\tan }^2}\theta } \right) + \tan \theta - 7 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + \tan \theta - 4 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + 4\tan \theta - 3\tan \theta - 4 = 0 \cr & \Rightarrow \tan \theta \left( {3\tan \theta + 4} \right) - 1\left( {3\tan \theta + 4} \right) = 0 \cr & \Rightarrow \left( {3\tan \theta + 4} \right)\left( {\tan \theta - 1} \right) = 0 \cr & \Rightarrow \tan \theta = 1\,\,\,\,\,\,\,\,\,\therefore \theta = {45^ \circ } \cr & \therefore \,\frac{{2\sin \theta + 3\cos \theta }}{{{\text{cosec}}\,\theta + \sec \theta }} \cr & = \frac{{2\left( {\frac{1}{{\sqrt 2 }}} \right) + 3\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 + \sqrt 2 }} \cr & = \frac{{\frac{5}{{\sqrt 2 }}}}{{2\sqrt 2 }} \cr & = \frac{5}{4} \cr} $$

Solution:

 cos(36° - A)cos(36° + A) + cos(54° - A)cos(54° + A) = cos(36° - A)cos(36° + A) + sin(36° + A)sin(36° - A) ∵ cos(A - B) = cosA.cosB + sinA.sinB = cos[36° - A - 36° - A] = cos(-2A) = cos2A

Solution:

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