Trigonometry Aptitude Questions for Competition Exams

(01). \ if\ sinA\ - \ cosA=\frac{\sqrt{3} -1}{2} \\ \\ than\ the\ value\ of\ sinA.cosA\ is\\\ \\ (a)\frac{1}{\sqrt{3}} \\\ \\ (b)\frac{\sqrt{3}}{2} \\\ \\ (c)\frac{1}{4} \\\ \\ (d)\frac{\sqrt{3}}{4}\\ \\ Read Solution

\Longrightarrow sinA-cosA=\frac{\sqrt{3} -1}{2}\\\ \\ ( square\ both\ sides )\\\ \\ \Longrightarrow ( sinA-cosA)^{2} =\left(\frac{\sqrt{3} -1}{2}\right)^{2}\\\ \\ \left( \ use\ formula\ of\ ( a-b)^{2} =a^{2} -2.a.b+b^{2}\right)\\\ \\ \Longrightarrow sin^{2} A-2.sinA.cosA+cos^{2} A=\frac{\left(\sqrt{3} -1\right)^{2}}{4}\\\ \\ \Longrightarrow sin^{2} A+cos^{2} A-2.sinA.cosA=\frac{\left(\sqrt{3}\right)^{2} -2.\sqrt{3} .1+( 1)^{2}}{4} \ \\\ \\ \left( \ we\ know\ sin^{2} \theta +cos^{2} \theta =1\right)\\\ \\ \Longrightarrow 1-2.sinA.cosA=\frac{3-2\sqrt{3} +1}{4}\\\ \\ \Longrightarrow 4( 1-2.sinA.cosA) =4-2\sqrt{3}\\\ \\ \Longrightarrow 4-8.sinA.cosA=4-2\sqrt{3}\\\ \\ \Longrightarrow 4-4+2\sqrt{3} =8.sinA.cosA\\\ \\ \Longrightarrow 2\sqrt{3} =8.sinA.cosA \\\ \\ \Longrightarrow sinA.cosA=\frac{2\sqrt{3}}{8}\\\ \\ \Longrightarrow sinA.cosA=\frac{\sqrt{3}}{4}\\\ \\

Option (d) is the right answer

(02). if\ \frac{sin\theta +cos\theta }{sin\theta -cos\theta } =\frac{5}{4} ,\\ \\ than\ the\ value\ of\ \frac{tan^{2} \theta -1}{tan^{2} \theta +1} \ is \\\ \\ (a)\frac{25}{16} \\\ \\ (b)\frac{41}{9} \\\ \\ (c)\frac{40}{41} \\\ \\ (d)\frac{41}{40}\\ \\ Read Solution

\Longrightarrow \frac{sin\theta +cos\theta }{sin\theta -cos\theta } =\frac{5}{4}\\\ \\ \Longrightarrow 4( sin\theta +cos\theta ) =5( sin\theta -cos\theta )\\\ \\ \Longrightarrow 4sin\theta +4cos\theta =5sin\theta -5cos\theta \\\ \\ \Longrightarrow 4sin\theta -5sin\theta =-5cos\theta -4cos\theta \\\ \\ \Longrightarrow -sin\theta =-9cos\theta \\\ \\ \Longrightarrow \frac{sin\theta }{cos\theta } =9 \\\ \\ \Longrightarrow tan\theta =9 \\\ \\ ( \ put\ the\ value\ tan\theta =9\ given\ equation)\\\ \\ =\ \frac{tan^{2} \theta -1}{tan^{2} \theta +1}\\\ \\ =\frac{9^{2} -1}{9^{2} +1}\\\ \\ =\frac{81-1}{81+1} \\\ \\ =\frac{80}{82}\\\ \\ =\frac{40}{41}\\\ \\

Option (c) is the right answer

(03). if\ tan15\degree =2-\sqrt{3} \ ,\ than\ the\ value\ of\\ \\ tan15\degree .cot75\degree +tan75\degree .cot15\degree \ is\\\ \\ (a) 14 \\\ \\ (b) 12 \\\ \\ (c) 10 \\\ \\ (d) 8 \\ \\ Read Solution

=\ tan15\degree .cot75\degree +tan75\degree .cot15\degree \\\ \\ =\ tan15\degree .cot( 90\degree -15\degree ) +tan( 90\degree -15\degree ) .cot15\degree \\\ \\ \\\ \\ \\\ \\ [ \ we\ know\ \ cot( 90\degree -\theta ) =tan\theta \ \ \&\ tan( 90\degree -\theta ) =cot\theta \ ]\\\ \\ =\ tan15\degree .tan15\degree +cot15\degree .cot15\degree \\\ \\ =tan^{2} 15\degree +cot^{2} 15\degree \ \ \ ..........\ equation\ no.1\\\ \\ \\\ \\ \ \ the\ value\ of\ cot15\degree \ is\\\ \\ \left( we\ know\ cot\theta =\frac{1}{tan\theta }\right)\\\ \\ so\ cot15\degree =\frac{1}{tan15\degree }\\\ \\ \left( \ put\ \ tan15\degree =2-\sqrt{3}\right)\\\ \\ \ cot15\degree =\frac{1}{tan15\degree }\\\ \\ =\frac{1}{2-\sqrt{3}}\\\ \\ =\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} \ \ \left[ multiply\ by\ \left( 2+\sqrt{3}\right)\right]\\\ \\ \left( \ we\ know\ ( a-b)( a+b) =a^{2} -b^{2}\right)\\\ \\ =\frac{2+\sqrt{3}}{( 2)^{2} -\left(\sqrt{3}\right)^{2}}\\\ \\ =\frac{2+\sqrt{3}}{4-3}\\\ \\ =2+\sqrt{3}\\\ \\ \\\ \\ \\\ \\ \left( \ put\ the\ value\ of\ tan\theta =2-\sqrt{3} \ \&\ cot\theta =2+\sqrt{3} \ in\ equation\ 1\right)\\\ \\ =tan^{2} 15\degree +cot^{2} 15\degree \\\ \\ =\left( 2-\sqrt{3}\right)^{2} +\left( 2+\sqrt{3}\right)^{2}\\\ \\ =\left[( 2)^{2} -2.2.\sqrt{3} +\left(\sqrt{3}\right)^{2}\right] +\left[( 2)^{2} +2.2.\sqrt{3} +\left(\sqrt{3}\right)^{2}\right]\\\ \\ =4-4\sqrt{3} +3+4-4\sqrt{3} +3\\\ \\ =8+6\\\ \\ =14\\\ \\

Option (a) is the right answer

(04) if\ A=tan11\degree .tan29\degree \\ \\ B=2\ cot61\degree .cot79\degree ,than;\\\ \\ (a) A=2B \\\ \\ (b) A=-2B \\\ \\ (c) 2A=B \\\ \\ (d) 2A=-B \\ \\ Read Solution

A=tan11\degree . tan29\degree \ \ \ .......equation\ no1\\\ \\ B=2\ cot61\degree .cot79\degree \ \ \ ........equation\ no2\\\ \\ equation\ 1\ is\ divided\ by\ equation\ 2\ \\\ \\ \frac{A}{B} =\frac{tan11\degree . tan29\degree }{2\ cot61\degree .cot79\degree }\\\ \\ \Longrightarrow \frac{A}{B} =\frac{tan11\degree . tan29\degree }{2\ cot( 90\degree -29\degree ) .cot( 90\degree -11\degree )}\\\ \\ [ we\ know\ cot( 90\degree -\theta ) =tan\theta ]\\\ \\ \Longrightarrow \frac{A}{B} =\frac{tan11\degree . tan29\degree }{2\ tan11\degree . tan29\degree }\\\ \\ \Longrightarrow \frac{A}{B} =\frac{1}{2}\\\ \\ 2A=B\\\ \\

Option (c) is the right answer

(05). if\ tan9\degree =\frac{p}{q} \\ \\ than\ the\ value\ of\ \frac{sec^{2} 81\degree }{1+cot^{2} 81\degree } \ is\\\ \\ (a)\frac{q}{p} \\\ \\ (b) 1 \\\ \\ (c)\frac{p^{2}}{q^{2}} \\\ \\ (d)\frac{q^{2}}{p^{2}}\\ \\ Read Solution


=\frac{sec^{2} 81\degree }{1+cot^{2} 81\degree }\\\ \\ \left( we\ know\ cosec^{2} \theta =1+cot^{2} \theta \right)\\\ \\ =\frac{sec^{2} 81\degree }{cosec^{2} 81\degree }\\\ \\ \left( we\ know\ sec\theta =\frac{1}{cos\theta } \ \&\ \ cosec\theta =\frac{1}{sin\theta }\right)\\\ \\ =\frac{\frac{1}{cos^{2} 81\degree }}{\frac{1}{sin^{2} 81\degree }}\\\ \\ =\frac{sin^{2} 81\degree }{cos^{2} 81\degree }\\\ \\ \left( we\ know\ tan\theta =\frac{sin\theta }{cos\theta }\right)\\\ \\ =tan^{2} 81\degree \\\ \\ =tan^{2}( 90\degree -9\degree )\\\ \\ \left[ we\ know\ tan( 90\degree -\theta ) =cot\theta \ \&\ cot\theta =\frac{1}{tan\theta }\right]\\\ \\ =cot^{2} 9\degree \\\ \\ =\frac{1}{tan^{2} 9\degree }\\\ \\ \left( put\ the\ value\ tan9\degree =\frac{p}{q}\right)\\\ \\ =\frac{1}{\left(\frac{p}{q}\right)^{2}}\\\ \\ =\frac{q^{2}}{p^{2}}\\\ \\

option (d) is the right answer

(06) if\ cos\theta +sec\theta =2\\ \\ \ the\ value\ of\ cos^{6} \theta +sec^{6} \theta \ is\ \\\ \\ (a) 4 \\\ \\ (b) 8 \\\ \\ (c) 1 \\\ \\ (d) 2 \\ \\ Read Solution

\Longrightarrow \ cos\theta +sec\theta =2 \\\ \\ \left( we\ know\ sec\theta =\frac{1}{cos\theta }\right)\\\ \\ \Longrightarrow cos\theta +\frac{1}{cos\theta } =2 \\\ \\ \Longrightarrow \frac{cos^{2} \theta +1}{cos\theta } =2 \\\ \\ \Longrightarrow cos^{2} \theta +1=2cso\theta \\\ \\ \Longrightarrow cos^{2} \theta +1-2cos\theta =0 \\\ \\ \left[ use\ formula\ ( a-b)^{2} =a^{2} +b^{2} -2.a.b\right] \\\ \\ \Longrightarrow ( cos \theta -1)^{2} =0 \\\ \\ \Longrightarrow cos\theta =1 \\\ \\ ( put\ the\ value\ cos\theta =1\ in\ given\ equation) \\\ \\ =cos^{6} \theta +sec^{6} \theta \\\ \\ \left( we\ know\ sec\theta =\frac{1}{cos\theta }\right)\\\ \\ =cos^{6} \theta +\frac{1}{cos^{6} \theta }\\\ \\ =( 1)^{6} +\frac{1}{( 1)^{6}} \\\ \\ =1+1 \\\ \\ =2 \\\ \\

Option (d) is the right answer

(07) the\ value\ of\\ \\ 152 ( sin30\degree +2\ cos^{2} 45\degree +3\ sin30\degree\\ \\ +4cos^{2} 45\degree +.......+17sin30\degree +18cos^{2} 45\degree ) \ is \\\ \\ (a) an\ integer\ but\ not\ perfect\ square\\\ \\ (b) a\ rational\ number\ but\ not\ an\ integer\\\ \\ (c) a\ perfect\ square\ of\ an\ integer \\\ \\ (d) irrational \\\ \\ Read Solution

=152( sin30\degree +2\ cos^{2} 45\degree +3\ sin30\degree +4cos^{2} 45\degree +.......\\ \\ \ \ \ \ \ \ \ \ +17sin30\degree +18cos^{2} 45\degree )\\\ \\ ( we\ know\ sin30\degree =\frac{1}{2} \ \&\ cos45\degree =\frac{1}{\sqrt{2}})\\\ \\ =152 [\frac{1}{2} +2\left(\frac{1}{\sqrt{2}}\right)^{2} +3\left(\frac{1}{2}\right) +4\left(\frac{1}{\sqrt{2}}\right)^{2} + \\ \\ \ \ \ \ \ \ \ \ \ \ ..........+17\left(\frac{1}{2}\right) +18\left(\frac{1}{\sqrt{2}}\right)^{2}]\\\ \\ =152 [\frac{1}{2} +2\left(\frac{1}{2}\right) +3\left(\frac{1}{2}\right) +4\left(\frac{1}{2}\right) +.........\\ \\ \ \ \ \ \ \ \ \ \ \ +17\left(\frac{1}{2}\right) +18\left(\frac{1}{2}\right) ] \\\ \\ =152\left(\frac{1}{2} +1+\frac{3}{2} +2+......+\frac{17}{2} +9\right) ....equation\ no1\\\ \\ it\ is\ sequence\ order\ in\ arithmetic\ progress( A.P)\\\ \\ A.P=\frac{n}{2}[ 2.a+( n-1) \times d]\\\ \\ =\frac{18}{2}\left[ 2\left(\frac{1}{2}\right) +( 18-1) \times \frac{1}{2}\right]\\\ \\ =9\left(\frac{19}{2}\right)\\\ \\ =\frac{171}{2}\\\ \\ ( \ put\ the\ value\ in\ equ.\ no1)\\\ \\ =152\times \frac{171}{2}\\\ \\ =76\times 171\\\ \\ =12,996 \\\ \\ =square\ root\ of\ 114\ \&\ it\ is\ integer\\\ \\

Option (c) is an integer

(08) if\ tan\theta +cot\theta =2\ ,than\ the\ value\ of\ \theta \ is\\\ \\ (a) 45\degree \\\ \\ (b) 60\degree \\\ \\ (c) 90\degree \\\ \\ (d) 30\degree \\ \\ Read Solution

\Longrightarrow tan\theta +cot\theta =2 \\\ \\ \left( we\ know\ cot\theta =\frac{1}{tan\theta }\right)\\\ \\ \Longrightarrow tan\theta +\frac{1}{tan\theta } =2\\\ \\ \Longrightarrow \frac{tan^{2} \theta +1}{tan\theta } =2\\\ \\ \Longrightarrow tan^{2} \theta +1=2tan\theta \\\ \\ \Longrightarrow tan^{2} \theta +1-2tan\theta =0 \\\ \\ \left[ use\ formula\ ( a-b)^{2} =a^{2} +b^{2} -2.a.b\right]\\\ \\ \Longrightarrow ( tan\theta -1)^{2} =0 \\\ \\ \Longrightarrow tan\theta -1=0\\\ \\ \Longrightarrow tan\theta =1\\\ \\ \Longrightarrow \theta =45\degree \\\ \\

Option (a) is the right answer

(09) The\ value\ of\\ \\ \frac{sin43\degree }{cos47\degree } +\frac{cos19\degree }{sin71\degree } -8\ cos^{2} 60\degree \ is\\\ \\ (a) 0 \\\ \\ (b) 1 \\\ \\ (c) 2 \\\ \\ (d) -1 \\ \\ Read Solution

=\ \frac{sin43\degree }{cos47\degree } +\frac{cos19\degree }{sin71\degree } -8\ cos^{2} 60\degree \\\ \\ =\ \frac{sin43\degree }{cos( 90\degree -43\degree )} +\frac{cos( 90\degree -71\degree )}{sin71\degree } -8\ cos^{2} 60\degree \\\ \\ \left[ we\ know\ cos( 90\degree -\theta ) =sin\theta \ \&\ cos60\degree =\frac{1}{2}\right]\\\ \\ =\ \frac{sin43\degree }{sin43\degree } +\frac{sin71\degree }{sin71\degree } -8\left(\frac{1}{2}\right)^{2}\\\ \\ =1+1-\frac{8}{4}\\\ \\ =2-2\\ =0\\\ \\

Option (a) is the right answer

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