Practice MCQ Questions and Answer on Trigonometry

Solution:

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

Solution:

 $$\eqalign{ & {\text{Arc length}} = \frac{\theta }{{180}} \times \pi r \cr & \Rightarrow 44 = \frac{{18}}{{180}} \times \frac{{22}}{7} \times r \cr & \Rightarrow r = 140\,{\text{m}} \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{\sec A - \tan A}}{{\sec A + \tan A}} \cr & = \frac{{\sec A - \tan A}}{{\sec A + \tan A}} \times \frac{{\sec A - \tan A}}{{\sec A - \tan A}} \cr & = \frac{{{{\left( {\sec A - \tan A} \right)}^2}}}{{{{\sec }^2}A - {{\tan }^2}A}} \cr & = \frac{{{{\sec }^2}A + {{\tan }^2}A - 2.\sec A.\tan A}}{1} \cr & = 1 + {\tan ^2}A + {\tan ^2}A - 2.\sec A.\tan A \cr & = 1 + 2{\tan ^2}A - 2.\sec A.\tan A \cr} $$

Solution:

 $$\eqalign{ & {\text{tan}}\theta - \cot \theta = 0 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {45^ \circ } \cr & {\text{tan }}{45^ \circ } - \cot {45^ \circ } = 0 \cr & 1 - 1 = 0 \cr & 0 - 0({\text{matched}}) \cr & So,\theta = {45^ \circ } \cr & \Rightarrow \sin \theta + \cos \theta \cr & \Rightarrow \sin {45^ \circ } + \cos {45^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \sqrt 2 \cr} $$

Solution:

 $$\eqalign{ & \frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ }\tan {{310}^ \circ }}} \cr & = \frac{{\tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right) - \tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}}{{1 + \tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right)\tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}} \cr & = \frac{{ - \tan {{40}^ \circ } + \cot {{40}^ \circ }}}{{1 + \tan {{40}^ \circ } \times \cot {{40}^ \circ }}} \cr & = \frac{{\cot {{40}^ \circ } - \tan {{40}^ \circ }}}{{1 + 1}} \cr & = \frac{{\frac{1}{\alpha } - \alpha }}{2} \cr & = \frac{{1 - {\alpha ^2}}}{{2\alpha }} \cr} $$

Solution:

 $$\eqalign{ & \frac{{\left( {\cos {9^ \circ } + \sin {{81}^ \circ }} \right)\left( {\sec {9^ \circ } + {\text{cosec}}\,{{81}^ \circ }} \right)}}{{\sin {{56}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }{\text{cosec}}\,{{65}^ \circ }}} \cr & = \frac{{\left( {\cos {9^ \circ } + \cos {9^ \circ }} \right)\left( {\sec {9^ \circ } + \sec {9^ \circ }} \right)}}{{\cos {{34}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }\sec {{25}^ \circ }}} \cr & = \frac{{2\cos {9^ \circ } \times 2\sec {9^ \circ }}}{{1 + 1}} \cr & = \frac{{4 \times \cos {9^ \circ }\sec {9^ \circ }}}{2} \cr & = 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\bf{Formula \,used:}} \cr & {\text{Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & {\text{Where, h = Hypotenuse, p = Perpendicular and b = base}} \cr & {\bf{Calculation:}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\text{Let, }}\tan A = x \cr & \Rightarrow {x^2} + 2x - 63 = 0 \cr & \Rightarrow {x^2} + 9x - 7x - 63 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 7\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 7} \right) \cr & \Rightarrow x + 9 = 0 \Rightarrow x = - 9\left[ {'' - ''{\text{ will be neglected because }}0 A \frac{\pi }{2}} \right] \cr & \Rightarrow x - 7 = 0 \Rightarrow x = 7 \cr & {\text{So,}}\tan A = \frac{7}{1} = \frac{p}{b} \cr & {\text{Using Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & h = \sqrt {{7^2} + {1^2}} \cr & h = \sqrt {50} \cr & {\text{So}},\,2\sin A + 5\cos A \cr & = \left[ {2 \times \frac{7}{{\sqrt {50} }}} \right] + \left[ {5 \times \frac{1}{{\sqrt {50} }}} \right] \cr & = \frac{{14}}{{\sqrt {50} }} + \frac{5}{{\sqrt {50} }} \cr & = \frac{{14 + 5}}{{\sqrt {50} }} \cr & = \frac{{19}}{{\sqrt {50} }} \cr & \therefore {\text{The value of}}\left( {2\sin A + 5\cos A} \right){\text{is }}\frac{{19}}{{\sqrt {50} }}. \cr} $$

Solution:

 $$\eqalign{ & r\sin \theta = 1,{\text{ }}r\cos \theta = \sqrt 3 \cr & {\text{Put }}\theta = {30^ \circ } \cr & r = 2 \cr & {\text{So}},{\text{ }}{r^2}\tan \theta \cr & = {\left( 2 \right)^2} \times {\text{tan}}{30^ \circ } \cr & = 4 \times \frac{1}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$

Solution:

 $$\eqalign{ & \sin A = \frac{5}{{13}} = \frac{P}{H};\,B = 12 \cr & \left[ {5,\,12,\,13\,{\text{are triplet}}} \right] \cr & 7\cot B = 24 \cr & \cot B = \frac{{24}}{7} = \frac{B}{P};\,H = 25 \cr & \left[ {7,\,24,\,25\,{\text{are triplet}}} \right] \cr & \Rightarrow \left( {\sec A\cos B} \right)\left( {{\text{cosec}}\,B\tan A} \right) \cr & = \left( {\frac{{13}}{{12}} \times \frac{{24}}{{25}}} \right)\left( {\frac{{25}}{7} \times \frac{5}{{12}}} \right) \cr & = \frac{{26}}{{25}} \times \left( {\frac{{25}}{7} \times \frac{5}{{12}}} \right) \cr & = \frac{{65}}{{42}} \cr} $$