# Trigonometry Aptitude Question

Question 01
If\ \theta \ be\ a\ positive\ acute\ angle\ satisfying \ cos^{2} \theta \ +\ cos^{4} \theta =1,\\ \\ \ \ \ \ \ then\ the\ value\ of\ tan^{2} \theta +tan^{4} \theta is \\\ \\ (a) \frac{3}{2} \ \ \ (b) 1 \ \ \ (c) \frac{1}{2} \ \ \ (d) 0

Given: cos^{2} \theta +cos^{4} \theta =1                 ----(i) \\ \\ cos^{4} \theta =1-cos^{2} \theta \\ \\ cos^{4} \theta =sin^{2} \theta \ \ \ \ \ \ \ \ \ \ \                   \left\{sin^{2} \theta +cos^{2} \theta =1\right\} \\ \\ cos^{2} \theta .cos^{2} \theta =sin^{2} \theta \\ \\ cos^{2} \theta =\frac{sin^{2} \theta }{cos^{2} \theta } \\ \\ cos^{2} \theta =tan^{2} \theta                 ----(ii) \\\ \\ Putting\ the\ value\ of\ equation (ii) \in\ equation (i), we\ get \\ \\ tan^{2} \theta +\left( tan^{2} \theta \right)^{2} =1 \\ \\ tan^{2} \theta +tan^{4} \theta =1 \\ \\ Ans. (b) 1

Question 02
If \ tan\theta =\frac{4}{3}, then\ the\ value\ of\ \frac{3sin\theta +2cos\theta }{3sin\theta -2cos\theta }\ is \\\ \\ (a) 0.5 \ \ \ (b) -0.5 \ \ \ (c) 3.0 \ \ \ (d) -3.0

\ Given: tan\theta =\frac{4}{3} \\ \\ \ To find: \frac{3sin\theta +2cos\theta }{3sin\theta -2cos\theta } \\\ \\ Dividing\ numerator \&\ denominator\ by\ cos\theta , we \ get \\\ \\     =   \frac{3tan\theta +2}{3tan\theta -2} \\\ \\   Putting\ the\ value\ of\ tan\theta \\ \\ =  \frac{3\times \frac{4}{3} +2}{3\times \frac{4}{3} -2} =\frac{6}{2} =3 \\\ \\ Ans. (c) 3

Question 03
The\ angles\ of\ a\ triangle\ are\ ( x+5) \degree , ( 2x-3) \degree and ( 3x+4) \degree.\\ \\ The\ value\ of\ x\ is \\\ \\ (a) 30 \ \ \ (b) 31 \ \ \ (c) 29 \ \ \ (d) 28

Angles \ of \ a \ triangle \ are \ ( x+5) \degree ,\ ( 2x-3) \degree ,\ ( 3x+4) \degree \\ \\ Sum\ of\ all\ angles\ in\ a\ triangle\ =180\degree \\ \\ ( x+5) \degree +( 2x-3) \degree +( 3x+4) \degree =180\degree \\ \\ 6x+6\degree =180\degree \\ \\  6( x+1) =180\degree \\ \\ x+1=30\degree \\ \\ x=29\degree \\ \\ Ans. (c) 29

Question 04
The\ expression\ \frac{tan57\degree +cot37\degree }{tan33\degree +cot53\degree }\ is\ equal\ to \\\ \\ (a) tan30\degree cot57\degree \\ \\ (b) tan57\degree cot37\degree \\ \\ (c) tan33\degree cot53\degree \\ \\ (d) tan33\degree cot37\degree

\frac{tan57\degree +cot37\degree }{tan33\degree +cot53\degree }\\\ \\ =\frac{tan( 90-33) \degree +cot( 90-53) \degree }{tan33\degree +cot53\degree } \\\ \\ = \frac{cot33\degree +tan53\degree }{tan33\degree +cot53\degree }                         \{tan( 90-\theta ) =cot\theta \}\\\ \\  = \frac{\frac{1}{tan33\degree } +tan53\degree }{tan33\degree +\frac{1}{tan53\degree }}                        \left\{cot\theta =\frac{1}{tan\theta }\right\} \\ \\  = \frac{1+tan53\degree .tan33\degree }{tan33\degree tan53\degree +1} \times \frac{tan53\degree }{tan33\degree }                    (Taking L.C.M.) \\\ \\ =  \frac{tan53\degree }{tan33\degree } \\\ \\ =  tan53\degree cot33\degree =tan( 90-37) \degree .cot( 90-57) \degree \\ \\ =cot37\degree tan57\degree \\\ \\Ans. (b) tan57\degree cot37\degree

Question 05
If \ sec^{2} \theta +tan^{2} \theta =\frac{7}{12},\\ \\ then\ sec^{4} \theta -tan^{4} \theta = ? \\\ \\ (a) \frac{7}{12} \ \ \ (b) \frac{1}{2} \ \ \ (c) \frac{1}{2} \ \ \ (d) 1

sec^{2} \theta +tan^{2} \theta =\frac{7}{12}\\\ \\ To find: sec^{4} \theta -tan^{4} \theta \\\ \\  =   \left( sec^{2} \theta +tan^{2} \theta \right)\left( sec^{2} \theta -tan^{2} \theta \right)    \left\{A^{2} -B^{2} =( A+B)( A-B)\right\} \\\ \\  =  \left( sec^{2} \theta +tan^{2} \theta \right) \times 1                        \left\{1+tan^{2} \theta =sec^{2} \theta \right\}\\\ \\= \frac{7}{12}                        (given) \\\ \\Ans. (a) \frac{7}{12}

Question 06
The\ value\ of \\\ \\ cos1\degree cos2\degree cos3\degree .......cos177\degree cos178\degree cos179\degree is; \\\ \\ (a) 0                        (b) \frac{1}{2}     (c) 1                        (d) \frac{1}{\sqrt{2}} \\\ \\ Read Solution

cos1\degree cos2\degree cos3\degree .....cos177\degree cos178\degree cos179\degree \\\ \\ We know, cos90\degree =0 \\\ \\ So, cos1\degree cos2\degree ...cos90\degree ...cos178\degree cos179\degree =0 \ \               [Multiplying \ series \ with \ 0 \ is \ 0] \\\ \\ Ans. (a) 0

Question 07
2cosec^{2} 23\degree cot^{2} 67\degree -sin^{2} 23\degree -sin^{2} 67\degree -cot^{2} 67\degree\\ \\ is \ equal\ to \\\ \\ (a) 1          (b) sec^{2} 23\degree    (c) tan^{2} 23\degree    (d) 0 \\\ \\ Read Solution

= 2cosec^{2} 23\degree cot^{2}( 90-23) \degree -sin^{2} 23\degree -sin^{2}( 90-23) \degree -cot^{2} 67\degree \\\ \\ =2cosec^{2} 23\degree tan^{2} 23\degree -sin^{2} 23\degree -cos^{2} 23\degree -cot^{2} 67\degree \\ \\ \ \ \ \ \ \{cot( 90-\theta ) =tan\theta ;\ sin( 90-\theta ) =cos\theta \} \\\ \\ \\\ \\ = 2\frac{1}{sin^{2} 23\degree }\frac{sin^{2} 23\degree }{cos^{2} 23\degree } -\left( sin^{2} 23\degree +cos^{2} 23\degree +cot^{2}( 90-23) \degree \right)\\\ \\                \left\{cosec\theta =\frac{1}{sin\theta }\right\} \\\ \\ \\\ \\ =  2\frac{1}{cos^{2} 23\degree } -\left( 1+tan^{2} 23\degree \right)\\ \\                              \left\{sin^{2} \theta +cos^{2} \theta =1 \right\} \\\ \\ =  2sec^{2} 23\degree -sec^{2} 23\degree \\\ \\ =  sec^{2} 23\degree \\\ \\ Ans. (b) sec^{2} 23\degree \\\ \\

Question 08
cos1\degree cos2\degree cos3\degree cos4\degree ....cos100\degree\\ \\ is \ equal \ to \\\ \\ (a) -1      (b) \frac{1}{4}     (c) 1      (d) 0 \\\ \\ Read Solution

To\ find:\ cos1\degree cos2\degree cos3\degree cos4\degree ....cos100\degree \\\ \\ We\ know,\ cos90\degree =\ 0 \\ \\ cos1\degree cos2\degree ....cos90\degree ...cos100\degree =0\                  [Multiplying\ series\ with\ 0\ is\ 0] \\ \\ Ans. (d) 0

Question 09
If \ \frac{sin\theta }{x} =\frac{cos\theta }{y}, \\ \\ then \ sin\theta -cos\theta \ is \ equal \ to \\\ \\ (a) x-y                    (b) x+y \\\ \\ (c) \frac{x-y}{\sqrt{x^{2} +y^{2}}}                  (d) \frac{y-x}{\sqrt{x^{2} +y^{2}}} \\\ \\ Read Solution

Let \ \frac{sin\theta }{x} =\frac{cos\theta }{y} =\frac{1}{k} \\\ \\ So,          x=k\ sin\theta               ----(i) \\ \\ y=k\ cos\theta           ----(ii)\\ \\ Squaring\ and\ adding\ equations\ (i) \& (ii), we get \\\ \\ x^{2} +y^{2} =k^{2} sin^{2} \theta +k^2 cos^{2} \theta \\\ \\ x^{2} +y^{2} =k^{2}\left( sin^{2} \theta +cos^{2} \theta \right) \\ \\ k=\sqrt{x^{2} +y^{2}}                 ----(iii) \\\ \\ therefore \ sin\theta -cos\theta =\frac{x}{k} -\frac{y}{k}             ---- from eq (i) \& (ii) \\\ \\ sin\theta -cos\theta =\frac{x-y}{k} =\frac{x-y}{\sqrt{x^{2} +y^{2}}}                          ---- value\ of\ k\ from\ eq (iii) \\\ \\ Ans. (c) \frac{x-y}{\sqrt{x^{2} +y^{2}}}

Question 10
If \ cos^{4} \theta -sin^{4} \theta =\frac{2}{3},\\ \\ then \ the \ value\ of\ 1-2sin^{2} \theta \ is \\\ \\ (a) \frac{4}{3}     (b) 0            (c) \frac{2}{3}     (d) \frac{1}{3} \\\ \\ Read Solution

So, cos^{4} \theta -sin^{4} \theta =\frac{2}{3} \\\ \\ \left( cos^{2} \theta +sin^{2} \theta \right)\left( cos^{2} \theta -sin^{2} \theta \right) =\frac{2}{3} \\ \\                   \left\{A^{2} -B^{2} =( A+B)( A-B)\right\} \\\ \\ 1\left( cos^{2} \theta -sin^{2} \theta \right) =\frac{2}{3}  \\ \\                                             \left\{sin^{2} \theta +cos^{2} \theta =1\right\} \\\ \\ 1-sin^{2} \theta -sin^{2} \theta =\frac{2}{3} \\ \\                                        \left\{cos^{2} \theta =1-sin^{2} \theta \right\} \\\ \\ 1-2sin^{2} \theta =\frac{2}{3} \\\ \\ Ans. (c)