# Trigonometry Quantitative Aptitude Problems

(01) the\ minimum\ value\ of\ \\ \\ sin^{2} \theta +cos^{2} \theta +sec^{2} \theta +cosec^{2} \theta +tan^{2} \theta +cot^{2} \theta \\

(a) 4
(b) 5
(c) 6
(d) 7

=\ \ sin^{2} \theta +cos^{2} \theta +sec^{2} \theta +cosec^{2} \theta +tan ^{2} \theta +cot^{2} \theta \\\ \\ we\ know\ that\ [sin^{2} \theta +cos^{2} \theta =1] \\ \\ also, \ [ tan^{2} \theta =sec^{2} \theta -1] \\ \\ [cot^{2} \theta =cosec^{2} \theta -1] \\\ \\ =\ 1+sec^{2} \theta +cosec^{2} \theta +\left( sec^{2} \theta -1\right) +\left( cosec^{2} \theta -1\right) \\ \\ \ \ \ \ \ \ \ \ \ \ \\\ \\ =\ 1+2sec^{2} \theta +2cosec^{2} \theta -2 \\\ \\ =\ 1+2\left( sec^{2} \theta +cosec^{2} \theta \right) -2 \\\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ =\ 2\left(\frac{1}{cos^{2} \theta } +\frac{1}{sin^{2} \theta }\right) -1\\\ \\ =\ 2\left(\frac{sin^{2} \theta +cos^{2} \theta }{cos^{2} \theta .sin^{2} \theta }\right) -1\\\ \\ =\ 2\left(\frac{1}{cos^{2} \theta .sin^{2} \theta }\right) -1\\\ \\ [ multiply\ by\ 4]\\\ \\ =\ 2\times 4\left(\frac{1}{4cos^{2} \theta .sin^{2} \theta }\right) -1\\\ \\ by\ sin\theta \ graph\\ \\ \ [ -1\leqslant sin^{2} 2 \theta \leqslant 1] \\\ \\ [0\leqslant sin^{2} 2 \theta \leqslant 1] \ \\\ \\ [ 0\geqslant \frac{1}{sin^{2} 2 \theta } \geqslant 1] \\\ \\ For\ min\ value\ Sin2 \theta=1\\ \\ =\ \frac{8}{sin^{2} 2 \theta } -1\\\ \\ = \ 8-1 \\\ \\ =\ 7 \\

Option (d) is the right answer

(02) if\ tan\theta =1,\ than\ the\ value\ of\\\ \\ \frac{8sin\theta +5cos\theta }{sin^{3} \theta -2cos^{3} \theta +7cos\theta }

(a) 1
(b) 2
(c) 3
(d) 4

tan\theta =1\\ \\ tan\theta =tan45\degree \\ \\ \theta =45\degree \\\ \\ put\ the\ value\ of\ given\ equation\\\ \\ =\ \frac{8sin\theta +5cos\theta }{sin^{3} \theta -2cos\theta +7cos\theta }\\\ \\ =\ \frac{8sin45\degree +5cos45\degree }{sin^{3} 45\degree -2cos^{3} 45\degree +7cos45\degree } \\\ \\ =\ \frac{8\left(\frac{1}{\sqrt{2}}\right) +5\left(\frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{\sqrt{2}}\right)^{3} -2\left(\frac{1}{\sqrt{2}}\right)^{3} +7\left(\frac{1}{\sqrt{2}}\right)}\\\ \\ =\ \ \frac{\frac{13}{\sqrt{2}}}{\frac{1-2+14}{2\sqrt{2}}}\\\ \\ =\ \frac{13}{\sqrt{2}} \times \frac{2\sqrt{2}}{13} \\\ \\ =\ 2\\

Option (b) is the right answer

(03) \ the\ value\ of\ \frac{sin\theta -2sin^{3} \theta }{2cos^{3} \theta \ -cos\theta } \ is\ equal\ to\\\ \\ (a) \ sin\theta \\\ \\ (b) cos\theta \\\ \\ (c) tan\theta \\\ \\ (d) cot\theta \\ Read Solution

=\frac{sin\theta -2sin^{3} \theta }{2cos^{3} \theta -cos\theta } \\\ \\ =\frac{sin\theta \left( 1-2sin^{2} \theta \right)}{cos\theta \left( 2cos^{2} \theta -1\right)}\\\ \\ \left[ we\ know\ sin^{2} \theta =1-cos^{2} \theta \ put\ the\ value\right]\\\ \\ =\frac{sin\theta }{cos\theta } \ \times \frac{\left[ 1-2\left( 1-cos^{2} \theta \right)\right]}{\left( 2cos^{2} \theta -1\right)}\\\ \\ =tan\theta \times \frac{\left( 1-2+2cos^{2} \theta \right)}{\left( 2cos^{2} \theta -1\right)}\\\ \\ =tan\theta \times \frac{\left( 2cos^{2} \theta -1\right)}{\left( 2cos^{2} \theta -1\right)}\\\ \\ =tan\theta
Option (c) is the right answer

(04) if\ r\ sin\theta =\frac{7}{2} \ and\ r\ cos\theta =\frac{7\sqrt{3}}{2} \\ \\ ,\ than\ the\ value\ of\ r\ is \\\ \\ (a) 4 \\ \\ (b) 3 \\ \\ (c) 5 \\ \\ (d) 7 \\ \\ Read Solution

r\ sin\theta =\frac{7}{2} \ \ \ \ \ \ \ \ \ \ ..........equation( 1)\\\ \\ r\ cos\theta =\frac{7\sqrt{3}}{2} \ \ \ \ \ ............equation( 2)\\\ \\ from\ eq.( 1) \ \&\ eq.\ ( 2) \ squaring\ both\ sides\\ \\ and\ adding\\\ \\ \Longrightarrow r^{2\ } sin^{2} \theta +r^{2} cos^{2} \theta =\left(\frac{7}{2}\right)^{2} +\left(\frac{7\sqrt{3}}{2}\right)^{2}\\\ \\ \left[ r^{2} \ is\ common\ from\ equation\right]\\\ \\ \Longrightarrow r^{2\ }\left( sin^{2} \theta + cos^{2} \theta \right) =\left(\frac{49}{4}\right) +\left(\frac{49\times 3}{4}\right)\\\ \\ \left[ we\ know\ sin^{2} \theta + cos^{2} \theta =1\right]\\\ \\ \Longrightarrow r^{2} \times 1=\frac{49}{4} +\frac{147}{4}\\\ \\ \Longrightarrow r^{2} =\frac{49+147}{4}\\\ \\ \Longrightarrow r^{2} =\frac{196}{4}\\\ \\ \Longrightarrow r^{2} =49\\\ \\ r=\sqrt{49}\\\ \\ r=7\\

option (d) is the right answer

(05) \ if\ tan\theta +cot\theta =5,\ than\ tan^{2} \theta +cot^{2} \theta \ is\\\ \\ (a) 23 \\ \\ (b) 24 \\ \\ (c) 25 \\ \\ (d) 26\\ Read Solution

\Longrightarrow tan\theta +cot\theta =5\\\ \\ [ squaring \ both\ sides]\\\ \\ \Longrightarrow ( tan\theta +cot\theta )^{2} =( 5)^{2}\\\ \\ \left[( a+b)^{2} =a^{2} +2ab+b^{2}\right]\\\ \\ \Longrightarrow tan^{2} \theta +2.tan\theta .cot\theta +cot^{2} \theta =25\\\ \\ [ we\ know\ tan\theta .cot\theta =1]\\\ \\ \Longrightarrow tan^{2} \theta +cot^{2} \theta +2=25\\\ \\ \Longrightarrow tan^{2} \theta +cot^{2} \theta =25-2\\ \ \\ \Longrightarrow tan^{2} \theta +cot^{2} \theta =23\\

Option (a) is the right answer

(06). \ the\ value\ of\ tan4\degree .tan43\degree .tan47\degree .tan86\degree \ is\\\ \\ (a) 2 \\ \\ (b) 3 \\ \\ (c) 1 \\ \\ (d) 4\\ Read Solution

=tan4\degree .tan43\degree .tan47\degree .tan86\degree \\\ \\ [ we\ know\ tan( 90\degree -\theta ) =cot\theta ]\\\ \\ =tan( 90\degree -86\degree ) .tan( 90\degree -47\degree ) .tan47\degree .tan86\degree \\\ \\ =cot86\degree .cot47\degree .tan47\degree .tan86\degree \\\ \\ [ we\ know\ tan\theta .cot\theta =1]\\\ \\ =cot86\degree .tan86\degree .cot47\degree .tan47\degree \\\ \\ =1\\

Option (c) is the right answer

(07) \ the\ numerical\ value\ of\ \frac{9}{cosec^{2} \theta } +4cos^{2} \theta +\frac{5}{1+tan^{2} \theta } \ is\\\ \\ (a) 7 \\ \\ (b) 9 \\ \\ (c) 4 \\ \\ (d) 5 \\ Read Solution

=\frac{9}{cosec^{2} \theta } +4cos^{2} \theta +\frac{5}{1+tan^{2} \theta }\\\ \\ \left[ we\ know\ sin\theta =\frac{1}{cosec\theta } \ \&\ \ sce^{2} \theta =1+tan^{2} \theta \right]\\\ \\ =9\ sin^{2} \theta +4cos^{2} \theta +5\frac{1}{sec^{2} \theta }\\\ \\ \left[ we\ know\ cos\theta =\frac{1}{sec\theta }\right]\\\ \\ =9\ sin^{2} \theta +4cos^{2} \theta +5cos^{2} \theta \\\ \\ =9\ sin^{2} \theta +9cos^{2} \theta \\\ \\ =9\ \left( sin^{2} \theta +cos^{2} \theta \right)\\\ \\ \left[ we\ know\ sin^{2} \theta +cos^{2} \theta =1\right]\\\ \\ =9\times 1\\\ \\ =9\\

option (b) is the right answer

(08) \ the\ value\ of\\ \\ tan1\degree .\ tan2\degree .\ tan3\degree ...........tan89\degree \ is\\ (a) \ 0 \\\ \\ (b) 1 \\\ \\ (c) \ \sqrt{3} \\\ \\ (d)\frac{1}{\sqrt{3}}\\ \\ Read Solution

=tan1\degree .\ tan2\degree .\ tan3\degree ...........tan89\degree \\\ \\ =( tan1\degree .\ tan89\degree ) \ ( \ tan2\degree .tan88\degree ) \ ( tan3\degree .tan87\degree ) \\ \\ ................ ( tan44\degree .tan46\degree ) \ tan45\degree \\\ \\ \\\ \\ =[ tan1\degree \ .tan( 90-1) \degree ] \ [ \ tan2\degree .tan( 90-2) \degree ] \ ...................[ tan44\degree .tan( 90-44) \degree ] \ tan45\degree \\\ \\ \\\ \\ [ we\ know\ tan( 90\degree -\theta ) =cot\theta ]\\\ \\ =( tan1\degree \ cot1\degree ) \ ( \ tan2\degree cot2\degree ) \ ...................\\ \\ \ \ \ \ ( tan44\degree cot.44\degree ) \ \ tan45\degree \\\ \\ ( we\ know\ tan\theta .cot\theta =1\ \&\ tan45\degree =1)\\\ \\ =1\times 1\times 1.........\times 1\times 1\\\ \\ =1\\

Option (b) is the right answer

(09) \ the\ value\ of \\ \\ sin^{2} 1\degree +sin^{2} 5\degree +sin^{2} 9\degree .......+sin^{2} 89\degree \ is\ \\\ \\ (a) 11\frac{1}{2} \\\ \\ (b) 11\sqrt{2} \\\ \\ (c) 11 \\\ \\ (d)\frac{11}{\sqrt{2}}\\ Read Solution

first\ identify\ the\ number\ of\ terms\\\ \\ 1+5+59+........+89\\\ \\ no.\ of\ term\ =\frac{Last\ number-First\ number}{difference\ of\ number} +1\\\ \\ =\frac{89-1}{4} +1\\\ \\ =23\ terms\\\ \\ =11\times 2+01\\\ \\ =11pairs+01\\\ \\ now\\\ \\ =sin^{2} 1\degree +sin^{2} 5\degree +sin^{2} 9\degree .......+sin^{2} 89\degree \\\ \\ =\left( sin^{2} 1\degree +sin^{2} 89\degree \right) +\left( sin^{2} 5\degree +sin^{2} 85\degree \right) +\left( sin^{2} 9\degree +sin^{2} 81\degree \right)\\ \\ .......+\left( sin^{2} 44\degree +sin^{2} 46\degree \right) +\ sin45\degree \\\ \\ \\\ \\ =\left[ sin^{2} 1+sin^{2}( 90-1) \degree \right] +\left[ sin^{2} 5+sin^{2}( 90-5) \degree \right] +\left[ sin^{2} 9+sin^{2}( 90-9) \degree \right] ............\\ \\ +\left[ sin^{2} 44\degree +sin^{2}( 90-44) \degree \right] +sin^{2} 45\degree \\\ \\ \\\ \\ [ \ sin( 90\degree -\theta ) =cos\theta ]\\\ \\ =\left( sin^{2} 1+cos^{2} 1\degree \right) +\left( sin^{2} 5+cos^{2} 5\degree \right) +\left( sin^{2} 9+cos^{2} 9\degree \right) ............\\ \\ +\left( sin^{2} 44\degree +cos^{2} 44\degree \right) +sin^{2} 45\degree \\\ \\ \\\ \\ \left[ we\ know\ sin^{2} \theta +cos^{2} \theta =1\ \&\ \ sin45\degree =\frac{1}{\sqrt{2}}\right]\\\ \\ =1+1+1............+up\ to\ 11\ pairs+01\ middle\ term\\\ \\ =1+1+1........+1+\left(\frac{1}{\sqrt{2}}\right)^{2}\\\ \\ =11+\left(\frac{1}{\sqrt{2}}\right)^{2}\\\ \\ =11+\frac{1}{2}\\\ \\ =\frac{22+1}{2}\\\ \\ =\frac{23}{21}\\\ \\ =11\frac{1}{2}\\\ \\

Option (a) is the right answer

(10) \ sin^{2} 5\degree +sin^{2} 6\degree +.......+sin^{2} 84\degree +sin^{2} 85\degree =?\\\ \\ (a) 30\frac{1}{2} \\\ \\ (b) 40\frac{1}{2} \\\ \\ (c) 40 \\\ \\ (d) 39\frac{1}{2}\\ Read Solution

first\ identify\ the\ number\ of\ term\\\ \\ 5+6+........+84+85\\\ \\ no.\ of\ terms\ =\frac{Last\ number-First\ number}{difference\ of\ number} +1\\\ \\ =\frac{85-5}{1} +1\\\ \\ =\frac{80}{1} +1 \\\ \\ =81\ term\\\ \\ =40\times 2+01\\\ \\ =40pairs+01\\\ \\ now\\\ \\ =sin^{2} 5\degree +sin^{2} 6\degree +.......+sin^{2} 84\degree +sin^{2} 85\degree \\\ \\ \\\ \\ =\left( sin^{2} 5\degree +sin^{2} 85\degree \right) +\left( sin^{2} 6\degree +sin^{2} 84\degree \right) +............\\ \\ \ \ \ \ \ \ +\left( sin^{2} 44\degree +sin^{2} 46\degree \right) +sin^{2} 45\degree \\\ \\ \\\ \\ =\left[ sin^{2} 5\degree +sin^{2}( 90-5) \degree \right] +\left[ sin^{2} 6\degree +sin^{2}( 90-6) \degree \right] +...............\\ \\ \ \ \ \ \ \ \ +\left[ sin^{2} 44\degree +sin^{2}( 90-44) \degree \right] +sin^{2} 45\degree \\\ \\ \\\ \\ [ we\ know\ sin( 90\degree -\theta ) =cos\theta ]\\\ \\ =\left( sin^{2} 5\degree +cos^{2} 5\degree \right) +\left( sin^{2} 6\degree +cos^{2} 6\degree \right) + \\ \\...............+\left( sin^{2} 44\degree +cos^{2} 44\degree \right)) +sin^{2} 45\degree \\\ \\ \\\ \\ =\left( sin^{2} 5\degree +cos^{2} 5\degree \right) +\left( sin^{2} 6\degree +cos^{2} 6\degree \right) + \\ \\ ...............+up\ to\ 40\ pairs+01\ middle\ term\\\ \\ \left[ we\ know\ sin^{2} \theta +cos^{2} \theta =1\ \&\ sin45\degree =\frac{1}{\sqrt{2}}\right]\\\ \\ =1+1+1+1+.....+1+\left(\frac{1}{\sqrt{2}}\right)^{2}\\\ \\ =40+\frac{1}{2}\\\ \\ =\frac{81}{2}\\\ \\ =40\frac{1}{2}\\\ \\

Option (b) is the right answer