Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & \sin \theta - \cos \theta = 0 \cr & \sin \theta = \cos \theta \cr & \theta = {45^ \circ } \cr & {\text{Then,}} \cr & sec\theta + {\text{cosec}}\theta \cr & = \sqrt 2 + \sqrt 2 \cr & = 2\sqrt 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{We know that }} \cr & {\text{tan}}\left( {{{90}^ \circ } - \theta } \right) = {\text{cot}}\theta \cr & {\text{and, cot}}\left( {{{90}^ \circ } - \theta } \right) = {\text{tan}}\theta \cr & \Rightarrow {\text{tan}}\left( {4\theta - {{50}^ \circ }} \right) = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow \cot \left[ {{{90}^ \circ } - \left( {4\theta - {{50}^ \circ }} \right)} \right] = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - \left( {4\theta - {{50}^ \circ }} \right) = \left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - 4\theta + {50^ \circ } = {50^ \circ } - \theta \cr & \Rightarrow {90^ \circ } = 3\theta \cr & {\text{then}},\theta = {30^ \circ } \cr} $$

Solution:

 (cosec a - sin a) (sec a - cos a) (tan a + cot a) Put a = 45° ⇒ (cosec45° - sin45°) (sec45° - cos45°) (tan45° + cot45°) $$\eqalign{ & \Rightarrow \left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)\left( {1 + 1} \right) \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \times 2 \cr & \Rightarrow \frac{1}{2} \times 2 \cr & \Rightarrow 1 \cr} $$

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:

 $$\eqalign{ & \sqrt {\frac{{\cot \theta + \cos \theta }}{{\cot \theta - \cos \theta }}} \cr & = \sqrt {\frac{{\cos \theta \left( {\frac{1}{{\sin \theta }} + 1} \right)}}{{\cos \theta \left( {\frac{1}{{\sin \theta }} - 1} \right)}}} \cr & = \sqrt {\frac{{1 + \sin \theta }}{{1 - \sin \theta }}} \cr & = \sqrt {\frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} \cr & = \frac{{1 + \sin \theta }}{{\cos \theta }} \cr & = \sec \theta + \tan \theta \cr} $$

Solution:

 $$\eqalign{ & \frac{{3\sin {{58}^ \circ }}}{{\cos {{32}^ \circ }}} + \frac{{3\sin {{42}^ \circ }}}{{\cos {{48}^ \circ }}} \cr & = \frac{{3\cos {{32}^ \circ }}}{{\cos {{32}^ \circ }}} + \frac{{3\cos {{48}^ \circ }}}{{\cos {{48}^ \circ }}} \cr & \left\{ {\therefore \,{{58}^ \circ } + {{32}^ \circ } = {{90}^ \circ }\,\& \,{{42}^ \circ } + {{48}^ \circ } = {{90}^ \circ }} \right\} \cr & = 3 + 3 \cr & = 6 \cr} $$

Solution:

 $$\eqalign{ & 2\sin \left( {\frac{{\pi x}}{2}} \right) = {x^2} + \frac{1}{{{x^2}}} \cr & {\text{Let }}x = 1 \cr & \Rightarrow 2\sin {90^ \circ } = {1^2} + \frac{1}{{{1^2}}} \cr & \Rightarrow 2 \times 1 = 1 + 1 \cr & 2 = 2\left( {{\text{Matched}}} \right) \cr & {\text{So, }}x = 1 \cr & \Rightarrow \left( {x - \frac{1}{x}} \right) \cr & \Rightarrow 1 - \frac{1}{1} \cr & \Rightarrow 0 \cr} $$

Solution:

 cosec(65° + θ) - sec(25° - θ) + tan220° - cosec270° = sec(25° - θ) - sec(25° - θ) + tan220° - sec220° = tan220° - sec220° = -1

Solution:

 Given, (sin25° + sin210° + sin215° + . . . . . . + sin285° ) + sin290° We know that sinθ = cos(90° - θ ) Therefore sin285° = cos2(90° - 85°) = cos25° Similary sin260° = cos2(90° - 60°) = cos240° And we also know that sin2θ + cos2θ = 1 There are 8 pair in given equation sin25° + cos25° = 1 sin210° + cos210° = 1 . . . . . . . . . . . . . . . . . . . . . . . . sin240° + cos240° = 1 i.e. 8 + sin245° + sin290° ⇒ 8 + $$\frac{1}{2}$$ + 1 = $$9\frac{1}{2}$$ Short Trick:  $$\left( {si{n^2}{5^ \circ }\, + \,{{\sin }^2}{{10}^ \circ }\, + \,{{\sin }^2}{{15}^ \circ }\, + \,.....\,{{\sin }^2}{{85}^ \circ }} \right)\, + $$         $$\,{\sin ^2}90$$ $$\eqalign{ & {\text{Number of terms}} \cr & = \left\{ {\left( {\frac{{85 - 5}}{5}} \right) + 1} \right\}{\text{ + 1}} \cr & {\text{ = }}\frac{{17}}{2}{\text{ + 1}} \cr & {\text{Sum of series}} = 9\frac{1}{2} \cr} $$

Solution:

 $$\eqalign{ & \Rightarrow \sin {\text{ }}7x = \cos 11x \cr & \Rightarrow 7x + 11x = {90^ \circ } \cr & \Rightarrow 18x = {90^ \circ } \cr & \Rightarrow x = {5^ \circ } \cr & \Rightarrow \tan 9x + \cot 9x \cr & \Rightarrow \tan {45^ \circ } + \cot {45^ \circ } \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$