# Trigonometry – Important Solved Aptitude Problems

01. if\ sin( 3x-20\degree )\ =\ cos( 3y+20\degree ) \ than\ the\ value\ of\ ( x+y) \ is \\\ \\ (a) 20\degree \\\ \\ (b) 30\degree \\\ \\ (c) 40\degree \\\ \\ (d) 45\degree \\\ \\ Read Solution

\Longrightarrow sin( 3x-20\degree ) =cos( 3y+20\degree ) \\\ \\ [ we\ know\ sin( 90\degree -\theta ) =cos\theta \ ] \\\ \\ \Longrightarrow sin( 3x-20\degree ) =sin[ 90\degree -( 3y+20\degree )] \\\ \\ \Longrightarrow sin( 3x-20\degree ) =sin[ 90\degree -3y-20\degree ]\\\ \\ \Longrightarrow sin( 3x-20\degree ) =sin( 70\degree -3y)\\\ \\ \Longrightarrow 3x-20\degree =70\degree -3y\\\ \\ \Longrightarrow 3x+3y=70\degree +20\degree \\\ \\ \Longrightarrow 3( x+y) =90\degree \\\ \\ \Longrightarrow x+y=\frac{90}{3} \\\ \\ \Longrightarrow x+y=30\degree \\\ \\

Option (b) is the right answer

02. \ if\ sec( 7\theta +28\degree ) =cosec( 30\degree -3\theta ) \ than\ the\ value\ of\ \theta \ is\ \\\ \\ (a) 8\degree \\\ \\ (b) 5\degree \\\ \\ (c) 60\degree \\\ \\ (d) 9\degree \\\ \\ Read Solution

Solution
\Longrightarrow sec( 7\theta +28\degree ) =cosec( 30\degree -3\theta )\\\ \\ [ \ we\ sec( 90\degree -\theta ) =cosec\theta ]\\\ \\ \Longrightarrow sec( 7\theta +28\degree ) =cosec[ 90\degree -( 30\degree -3\theta )] \\\ \\ \Longrightarrow sec( 7\theta +28\degree ) =sec( 90\degree -30\degree +3\theta )\\\ \\ \Longrightarrow sec( 7\theta +28\degree ) =sec( 60\degree +3\theta )\\\ \\ \Longrightarrow 7\theta +28\degree =60\degree +3\theta \\\ \\ \Longrightarrow 7\theta -3\theta =60\degree -28\degree \\\ \\ \Longrightarrow 4\theta =32\degree \\\ \\ \Longrightarrow \theta =\frac{32}{4} \\\ \\ \theta =8\degree \\\ \\

Option (a) is the right answer

03.\ if\ tan7\theta tan2\theta =1,\ than\ the\ value\ of\ tan3\theta \ is \\\ \\ (a) \ \sqrt{3} \\\ \\ (b) -\frac{1}{\sqrt{3}} \\\ \\ (c) \ \frac{1}{\sqrt{3}} \\\ \\ (d) -\sqrt{3}\\\ \\ Read Solution

\Longrightarrow tan7\theta tan2\theta =1\\\ \\ \Longrightarrow tan7\theta =\frac{1}{tan2\theta }\\\ \\ \Longrightarrow tan7\theta =cot2\theta \\\ \\ [ we\ know\ tan( 90\degree -\theta ) =cot\theta ]\\\ \\ \Longrightarrow tan7\theta =tan( 90\degree -2\theta )\\\ \\ \Longrightarrow 7\theta =90\degree -2\theta \\\ \\ \Longrightarrow 7\theta +2\theta =90\degree \\\ \\ \Longrightarrow 9\theta =90\degree \\\ \\ \Longrightarrow \theta =\frac{90}{9}\\\ \\ \theta =10 \\\ \\ now\\\ \\ =tan3\theta \\\ \\ ( put\ the\ value\ of\ \theta =10\degree ) \\\ \\ =tan( 3\times 10) \\\ \\ =tan30\degree \\\ \\ =\frac{1}{\sqrt{3}}\\\ \\

Option (c) is the right answer

04. Evaluate\ 3\ cos80\degree \ cosec10\degree +2\ cos59\degree \ cosec31\degree \\\ \\ (a) 1 \\\ \\ (b) 3 \\\ \\ (c) 2 \\\ \\ (d) 5\\\ \\ Read Solution

=3\ cos80\degree \ cosec10\degree +2\ cos59\degree \ cosec31\degree \\\ \\ \left( we\ know\ cosec\theta =\frac{1}{sin\theta }\right)\\\ \\ =3\ cos80\degree \ \frac{1}{sin10\degree } +2\ cos59\degree \ \frac{1}{sin31\degree }\\\ \\ [ we\ know\ sin( 90\degree -\theta ) =cos\theta ]\\\ \\ =3\ cos80\degree \ \frac{1}{sin( 90\degree -80\degree )} +2\ cos59\degree \ \frac{1}{sin( 90\degree -59\degree )}\\\ \\ [ we\ know\ sin( 90\degree -\theta ) =cos\theta ] \\\ \\ =3\ cos80\degree \ \frac{1}{cos80\degree } +2\ cos59\degree \ \frac{1}{cos59\degree }\\\ \\ =3+2\\\ \\ =5\\\ \\

Option (d) is the right answer

05. If\ sin7x=cos11x,\ than\ the\ value\ of\ tan9x+cot9x\ is\\\ \\ (a) 1 \\\ \\ (b) 2 \\\ \\ (c) 3 \\\ \\ (d) 4 \\\ \\ Read Solution

\Longrightarrow sin7x=cos11x \\\ \\ [ we\ know\ cos( 90\degree -\theta ) =sin\theta ]\\\ \\ \Longrightarrow sin7x=sin( 90\degree -11x)\\\ \\ \Longrightarrow 7x=90\degree -11x \\\ \\ \Longrightarrow 7x+11x=90\degree \\\ \\ \Longrightarrow 18x=90\degree \\\ \\ \Longrightarrow x=\frac{90}{18} \\\ \\ x=5\degree \\\ \\ now \\\ \\ =tan9x+cot9x\ \\\ \\ ( put\ the\ value\ of\ x=5\degree )\\\ \\ =tan( 9\times 5) +cot( 9\times 5) \\\ \\ =tan45\degree +cot45\degree \\\ \\ ( we\ know\ tan45\degree =1\ \ \ cot45\degree =1) \\\ \\ =1+1\\\ \\ =2\\

Option (b) is the right answer

06. \ the\ value\ of\ sin^{2} 65\degree +sin^{2} 25\degree +cos^{2} 35\degree +cos^{2} 55\degree \ is\ \\\ \\ (a) 0 \\\ \\ (b) 1 \\\ \\ (c) 2 \\\ \\ (d)\frac{1}{2}\\\ \\ Read Solution

=sin^{2} 65\degree +sin^{2} 25\degree +cos^{2} 35\degree +cos^{2} 55\degree \\\ \\ =sin^{2} 65\degree +sin^{2}( 90\degree -65\degree ) +cos^{2} 35\degree +cos^{2}( 90\degree -35\degree )\\\ \\ [ we\ know\ sin( 90\degree -\theta ) =cos\theta \ \&\ cos( 90\degree -\theta ) =sin\theta ]\\\ \\ =sin^{2} 65\degree +cos^{2} 65\degree +cos^{2} 35\degree +sin^{2} 35\degree \\\ \\ \left( we\ know\ sin^{2} \theta +cos^{2} \theta =1\right)\\\ \\ =1+1\\\ \\ =2\\\ \\

Option (C) is the right answer

07. if\ sin3A=cos( A-26\degree ) ,\ where\ 3A\ is\ acute\ angle\ than\ the\ value\ of\ A\ is\\\ \\ (a) 29\degree \\\ \\ (b) 26\degree \\\ \\ (c) 23\degree \\\ \\ (d) 28\degree \\\ \\ Read Solution

\Longrightarrow \ sin3A=cos( A-26\degree )\\\ \\ \Longrightarrow \ sin3A=sin[ 90\degree -( A-26\degree )]\\\ \\ \Longrightarrow sin3A=sin( 90\degree -A+26\degree )\\\ \\ \Longrightarrow 3A=116\degree -A\\\ \\ \Longrightarrow 3A+A=116\degree \\\ \\ \Longrightarrow 4A=116\degree \\\ \\ \Longrightarrow A=\frac{116}{4}\\\ \\ \Longrightarrow A=29\degree \\\ \\

Option (a) is the right answer

08. the\ value\ of\ sin^{2} 22\degree +sin^{2} 68\degree +cot^{2} 30\degree \ is\\\ \\ (a)\frac{5}{3} \\\ \\ (b)\frac{3}{4} \\\ \\ (c) 3 \\\ \\ (d) 4 \\\ \\ Read Solution

=\ sin^{2} 22\degree +sin^{2} 68\degree +cot^{2} 30\degree \\\ \\ =\ sin^{2} 22\degree +sin^{2}( 90\degree -22\degree ) +cot^{2} 30\degree \\\ \\ =\ sin^{2} 22\degree +cos^{2} 22\degree +cot^{2} 30\degree \\\ \\ =1+\left(\sqrt{3}\right)\\\ \\ =1+3\\\ \\ =4\\\ \\

Option (d) is the right answer