Practice MCQ Questions and Answer on Trigonometry

Solution:

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

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:

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr & \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$

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:

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

Solution:

 $$\eqalign{ & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & \Rightarrow {\text{se}}{{\text{c}}^2}\alpha - 1 = 1 + 2\left( {{\text{se}}{{\text{c}}^2}\beta - 1} \right) \cr & \Rightarrow {\sec ^2}\alpha - 1 = 2{\sec ^2}\beta - 1 \cr & \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}\alpha }} = \frac{2}{{{\text{co}}{{\text{s}}^2}\beta }} \cr & \Rightarrow \sqrt 2 {\text{cos}}\alpha = {\text{cos}}\beta \cr & \therefore \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & {\text{Put }}\beta = {45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2.{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 3 \cr & {\text{ tan }}\alpha = \sqrt 3 \cr & \alpha = {60^ \circ } \cr & {\text{Put }}\alpha = {60^ \circ },{\text{and }}\beta = {45^ \circ } \cr & = \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta \cr & = \sqrt 2 {\text{cos }}{60^ \circ } - {\text{cos }}{45^ \circ } \cr & = \sqrt 2 \times \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & \frac{{\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta } \right)\left( {\cot \theta + 1} \right)\cos \theta }}{{\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)\left( {1 + \tan \theta } \right){\text{cosec}}\,\theta }} - 1 \cr & {\text{Put }}\theta = {45^ \circ } \cr & = \frac{{\left( {1 - 2 \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right)\left( {1 + 1} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4}} \right)\left( {1 + 1} \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {1 - 2 \times \frac{1}{4}} \right)\left( 2 \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{1}{4} + \frac{1}{4}} \right)\left( {1 + 1} \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {\frac{1}{2}} \right)\left( 2 \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{1}{2}} \right)\left( 2 \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 }} - 1 \cr & = \frac{1}{2} - 1 \cr & = - \frac{1}{2} \cr & {\text{Option C is answer}} \cr & \Rightarrow - {\sin ^2}\theta \cr & = - {\sin ^2}{45^ \circ } \cr & = - {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & = - \frac{1}{2} \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - A} \right) + \cot \left( {{{90}^ \circ } - A} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2A} \right)} \right]}} \cr & {\text{By putting }}A = {45^ \circ } \cr & \Rightarrow \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right) + \cot \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2 \times {{45}^ \circ }} \right)} \right]}} \cr & \Rightarrow \frac{{{{\left[ {\tan {{45}^ \circ } + \cot {{45}^ \circ }} \right]}^2}}}{{2{{\sec }^2}\left( {{{90}^ \circ } - {{90}^ \circ }} \right)}} \cr & \Rightarrow \frac{{{{\left[ {1 + 1} \right]}^2}}}{{2{{\sec }^2}{0^ \circ }}} \cr & \Rightarrow \frac{4}{{2 \times 1}} \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & \alpha + \beta = {90^ \circ } \cr & {\text{and }}\alpha :\beta = 2:1 \cr & 2x + x = {90^ \circ } \cr & x = {30^ \circ } \cr & {\text{ }}\alpha = {60^ \circ } \cr & \beta = {30^ \circ } \cr & \frac{{{\text{cos }}\alpha }}{{\cos \beta }} \cr & = \frac{{\cos {{60}^ \circ }}}{{\cos {{30}^ \circ }}} \cr & = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} \cr & = \frac{1}{2} \times \frac{2}{{\sqrt 3 }} \cr & = \frac{1}{{\sqrt 3 }} \cr & = 1:\sqrt 3 \cr} $$

Solution:

 cos4α - sin4α = $$\frac{5}{6}$$ (cos2α + sin2α)(cos2α - sin2α) = $$\frac{5}{6}$$ 1 × (cos2α - 1 + cos2α) = $$\frac{5}{6}$$ 2cos2α - 1 = $$\frac{5}{6}$$