Practice MCQ Questions and Answer on Trigonometry

Solution:

 6tanA(tanA + 1) = 5 - tanA 6tan2A + 6tanA + tanA = 5 6tan2A + 7tanA = 5 6tan2A + 7tanA - 5 = 0 6tan2A + 10tanA - 3tanA - 5 = 0 2tanA(3tanA + 5) - 1(3tanA + 5) = 0 (2tanA - 1)(3tanA + 5) = 0 tanA = $$\frac{1}{2}$$ and tanA = $$ - \frac{5}{2}$$ taking tanA $$ = \frac{1}{2} = \frac{P}{B}$$ H = √5 $$\eqalign{ & \therefore \sin A + \cos A \cr & = \frac{P}{H} + \frac{B}{H} \cr & = \frac{1}{{\sqrt 5 }} + \frac{2}{{\sqrt 5 }} \cr & = \frac{3}{{\sqrt 5 }} \cr} $$

Solution:

 $$\eqalign{ & {\text{In }}\vartriangle {\text{ABC sin2}}{{\text{1}}^ \circ }{\text{ = }}\frac{x}{y} \cr & {\text{AB}} = x \cr & {\text{AC}} = y \cr & {\text{BC}} = \sqrt {{y^2} - {x^2}} \cr & \Rightarrow {\text{sec2}}{{\text{1}}^ \circ } - \sin {69^ \circ } \cr & \Rightarrow \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & \Rightarrow \frac{{{{\left( {{\text{AC}}} \right)}^2} - {{\left( {{\text{BC}}} \right)}^2}}}{{\left( {{\text{BC}}} \right)\left( {{\text{AC}}} \right)}} = \frac{{{y^2} - {{\left( {\sqrt {\left( {{y^2} - {x^2}} \right)} } \right)}^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr & \Rightarrow \frac{{{y^2} - {y^2} + {x^2}}}{{y\sqrt {{y^2} + {x^2}} }} = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$

Solution:

 $$\eqalign{ & \frac{{\left( {\cos {9^ \circ } + \sin {{81}^ \circ }} \right)\left( {\sec {9^ \circ } + {\text{cosec}}\,{\text{8}}{{\text{1}}^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}71 + {{\cos }^2}{{15}^ \circ } - {{\tan }^2}{{19}^ \circ } + {{\cos }^2}{{75}^ \circ }}} \cr & = \frac{{\left( {\cos {9^ \circ } + \cos {9^ \circ }} \right)\left( {\sec {9^ \circ } + \sec {9^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}71 + {{\cos }^2}{{15}^ \circ } - {{\cot }^2}{{71}^ \circ } + {{\sin }^2}{{15}^ \circ }}} \cr & = \frac{{\left( {2\cos {9^ \circ }} \right)\left( {2\sec {9^ \circ }} \right)}}{{1 + 1}} \cr & = 2 \cr} $$

Solution:

 $${\text{152}}\left( {\sin {{30}^ \circ }\, + \,2{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,3\sin {{30}^ \circ }\, + \,4{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,.....\, + \,17\sin {{30}^ \circ }\, + \,18{\text{co}}{{\text{s}}^2}{{45}^ \circ }} \right)$$ $$ = 152\left\{ {\frac{1}{2}\, + \,2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}\, + \,3\, \times \,\frac{1}{2}\, + \,.....\,\, + \,17 \times \frac{1}{2}\, + \,18{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right\}$$ $$\eqalign{ & = 152\left\{ {\frac{1}{2}\, + \,1\, + \,1\frac{1}{2} + \,.....\,8\frac{1}{2}\, + \,9} \right\} \cr & = {\text{This is in A}}{\text{.P}}{\text{. where}} \cr & a = \frac{1}{2},{\text{ }}d = \frac{1}{2},{\text{ }}n = 18 \cr & = 152\left\{ {\frac{{18}}{2}\left( {2\, \times \,\frac{1}{2}\, + \,\left( {18\, - \,1} \right)\frac{1}{2}} \right)} \right\} \cr & = 152\left\{ {\frac{{18}}{2}\left( {1\, + \,\frac{{17}}{2}} \right)} \right\} \cr & = 152 \times 9 \times \frac{{19}}{2} \cr & = 12996 \cr & = \sqrt {12996} \cr & = 114 \cr} $$

Solution:

 $$\eqalign{ & \sec \theta + \tan \theta = p \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{p}{{\frac{1}{p}}} \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{{{p^2}}}{1} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{{\sec \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{\sec \theta .\cos \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{{\text{cosec }}\theta }}{1} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & {\text{Apply again componendo and dividendo}} \cr & \frac{{{\text{cosec }}\theta + 1}}{{{\text{cosec }}\theta - 1}} = \frac{{{p^2}}}{1} \cr} $$

Solution:

Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board

Solution:

 $$\eqalign{ & 1 + 2{\tan ^2}\theta + 2\sin \theta {\sec ^2}\theta = \frac{a}{b} \cr & 1 + {\tan ^2}\theta + {\tan ^2}\theta + 2\sin \theta \frac{1}{{{{\cos }^2}\theta }} = \frac{a}{b} \cr & {\sec ^2}\theta + {\tan ^2}\theta + 2\tan \theta \sec \theta = \frac{a}{b} \cr & {\left( {\sec \theta + \tan \theta } \right)^2} = \frac{a}{b} \cr & {\left( {\frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & {\left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & \frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 + {{\sin }^2}\theta }} = \frac{a}{b} \cr & \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{a}{b} \cr & {\text{Use componendo and dividendo}} \cr & \frac{1}{{\sin \theta }} = {\text{cosec}}\,\theta = \frac{{a + b}}{{a - b}} \cr} $$

Solution:

 $$\eqalign{ & \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & A = {60^ \circ },\,\,x = {15^ \circ }{\text{ given,}} \cr & \Rightarrow \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{60}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}} = \frac{a}{b} \cr & \Rightarrow \frac{{\cos {{75}^ \circ }}}{{\cos {{135}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{\sin {{15}^ \circ }}}{{ - \sin {{45}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{{{{\sin }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = 2{\sin ^2}{15^ \circ }........\left( {\text{i}} \right) \cr & \Rightarrow \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{180}^ \circ }} \right)}} = \frac{b}{c} \cr & \Rightarrow \frac{{ - \sin {{45}^ \circ }}}{{ - \cos {{15}^ \circ }}} = \frac{b}{c} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = \frac{{{{\cos }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = 2{\cos ^2}{15^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{Now, }}{\left( {\frac{a}{b} + \frac{c}{b}} \right)^2} + {\left( {\frac{a}{b} - \frac{c}{b}} \right)^2} \cr & = 2\left[ {\frac{{{a^2}}}{{{b^2}}} + \frac{{{c^2}}}{{{b^2}}}} \right] \cr & = 2\left[ {2{{\sin }^2}{{15}^ \circ } + 2{{\cos }^2}{{15}^ \circ }} \right] \cr & = 4 \cr} $$

Solution:

 $$\eqalign{ & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {60^ \circ }{\text{option A}} \cr & \Rightarrow 2{\sin ^2}{60^ \circ } = 3\cos {60^ \circ } \cr & \Rightarrow 2{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = 3\left( {\frac{1}{2}} \right) \cr & \Rightarrow \frac{3}{2} = \frac{3}{2}{\text{ }}\left( {{\text{LHS}} = {\text{RHS}}} \right) \cr} $$

Solution:

 $$\eqalign{ & {\text{2cos}}\theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & {\text{When,}} \cr & ax \mp by = m \cr & {\text{then, }}bx \mp ay = \sqrt {{a^2} + {b^2} - {m^2}} \cr & 2\cos \theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \cos \theta + 2\sin \theta = \sqrt {4 + 1 - \frac{1}{2}} \cr & \Rightarrow \cos \theta + 2\sin \theta = \frac{3}{{\sqrt 2 }} \cr} $$