Practice MCQs for JEE Mathematics Trigonometric Functions
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Multiple Choice Questions
(a) \(0\)
(b) \(-1\)
(c) \(1\)
(d) \(2\)
Show Answer & Explanation
Answer: (c) \(1\)
Explanation:
1. Pair \(\cot\theta\cot(90^\circ-\theta)=\cot\theta\tan\theta=1\); the middle term is \(\cot45^\circ=1\).
2. So the product is 1.
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(1\)
(b) \(2\)
(c) \(\frac12\)
(d) \(0\)
Show Answer & Explanation
Answer: (d) \(0\)
Explanation:
1. \(\sin(360^\circ-\theta)=-\sin\theta\), so terms pair off to zero (and \(\sin180^\circ=\sin360^\circ=0\)).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \( rac\pi4\)
(b) \(\pi\)
(c) \( rac\pi6\)
(d) \( rac\pi2\)
Show Answer & Explanation
Answer: (a) \( rac\pi4\)
Explanation:
1. \(\tan(A+B)=\dfrac{\frac12+\frac13}{1-\frac16}=1\), so \(A+B=\dfrac\pi4\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(120^\circ,360^\circ\)
(b) \(120^\circ,240^\circ\)
(c) \(240^\circ,90^\circ\)
(d) \(60^\circ,270^\circ\)
Show Answer & Explanation
Answer: (b) \(120^\circ,240^\circ\)
Explanation:
1. Cosine is negative in quadrants II and III with reference angle \(60^\circ\): \(120^\circ\) and \(240^\circ\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(2n\pi+\frac\pi6\)
(b) \(n\pi+\frac\pi6\)
(c) \(2n\pi-\frac\pi6\)
(d) \(n\pi-\frac\pi6,\ n\in I\)
Show Answer & Explanation
Answer: (d) \(n\pi-\frac\pi6,\ n\in I\)
Explanation:
1. \(\tan\theta=\tan\left(-\frac\pi6\right)\), so \(\theta=n\pi-\frac\pi6\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(n\pi\pm\frac\pi4\)
(b) \(n\pi\pm\frac\pi6\)
(c) \(2n\pi+\frac\pi4\)
(d) \(2n\pi\pm\frac\pi6\)
Show Answer & Explanation
Answer: (a) \(n\pi\pm\frac\pi4\)
Explanation:
1. \(2\tan^2\theta=1+\tan^2\theta\) gives \(\tan^2\theta=1\), so \(\theta=n\pi\pm\frac\pi4\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\frac y3\)
(b) \(2y\)
(c) \(y\)
(d) \(\frac32y\)
Show Answer & Explanation
Answer: (c) \(y\)
Explanation:
1. Multiply numerator and denominator of \(y\) by \((1+\sin\alpha-\cos\alpha)\): denominator becomes \((1+\sin\alpha)^2-\cos^2\alpha=2\sin\alpha(1+\sin\alpha)\).
2. So \(y=\dfrac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(3\)
(b) \(0\)
(c) \(1\)
(d) \(2\)
Show Answer & Explanation
Answer: (d) \(2\)
Explanation:
1. \(\sin x+1=2\cos^2x=2-2\sin^2x\) gives \(2\sin^2x+\sin x-1=0\), so \(\sin x=\frac12\) or \(-1\).
2. \(\sin x=-1\) makes \(\cos x=0\) (undefined), rejected. \(\sin x=\frac12\) gives \(x=\frac\pi6,\frac{5\pi}6\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(1\)
(b) \(0\)
(c) \(\frac{\sqrt3}2\)
(d) \(-1\)
Show Answer & Explanation
Answer: (a) \(1\)
Explanation:
1. \(\tan15^\circ\tan75^\circ=\tan15^\circ\cot15^\circ=1\) and \(\tan45^\circ=1\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\frac38\)
(b) \(\frac34\)
(c) \(\frac58\)
(d) \( rac18\)
Show Answer & Explanation
Answer: (d) \( rac18\)
Explanation:
1. \(\cos\frac{7\pi}8=-\cos\frac\pi8\) and \(\cos\frac{5\pi}8=-\cos\frac{3\pi}8\).
2. Product \(=\sin^2\frac\pi8\sin^2\frac{3\pi}8=\left(\sin\frac\pi8\cos\frac\pi8\right)^2=\left(\frac12\sin\frac\pi4\right)^2=\frac18\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\frac7{25}\)
(b) \(\frac6{25}\)
(c) \(\frac8{25}\)
(d) \(\frac3{25}\)
Show Answer & Explanation
Answer: (a) \(\frac7{25}\)
Explanation:
1. \(\theta=\dfrac lr=\dfrac{21}{75}=\dfrac7{25}\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) 1.50 cm
(b) 1.57 cm
(c) 1.35 cm
(d) 1.20 cm
Show Answer & Explanation
Answer: (b) 1.57 cm
Explanation:
1. \(l=r\theta=3\times\frac{3.14}{6}=1.57\) cm.
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(50^\circ\)
(b) \(210^\circ\)
(c) \(100^\circ\)
(d) \(60^\circ\)
Show Answer & Explanation
Answer: (b) \(210^\circ\)
Explanation:
1. Length of wire \(=2\pi\times7=14\pi\).
2. \(\theta=\dfrac{14\pi}{12}=\dfrac{7\pi}6=210^\circ\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) 37.5 cm
(b) 32.8 cm
(c) 34.5 cm
(d) 35.7 cm
Show Answer & Explanation
Answer: (d) 35.7 cm
Explanation:
1. \(r=\dfrac{l}{\theta}=\dfrac{37.4}{\pi/3}=\dfrac{37.4\times3\times7}{22}=35.7\) cm.
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(12^\circ36'\)
(b) \(12^\circ30'\)
(c) \(11^\circ36'\)
(d) \(11^\circ12'\)
Show Answer & Explanation
Answer: (a) \(12^\circ36'\)
Explanation:
1. \(\theta=\frac{22}{100}\) rad \(=0.22\times\frac{180\times7}{22}=12.6^\circ=12^\circ36'\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\frac1{\sqrt{10}}\)
(b) \(\frac2{\sqrt{10}}\)
(c) \(\frac{-3}{\sqrt{10}}\)
(d) \(\frac{-5}{\sqrt{10}}\)
Show Answer & Explanation
Answer: (c) \(\frac{-3}{\sqrt{10}}\)
Explanation:
1. \(\sec^2\theta=10\), \(\cos\theta=-\frac1{\sqrt{10}}\) in quadrant III.
2. \(\sin\theta=\tan\theta\cos\theta=-\frac3{\sqrt{10}}\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\cot\theta\)
(b) \(-\tan\theta\)
(c) \(-\cot\theta\)
(d) \(\tan\theta\)
Show Answer & Explanation
Answer: (d) \(\tan\theta\)
Explanation:
1. \(\sec\theta-x\cos\theta\tan\theta=\cos\theta\), i.e. \(x\sin\theta=\sec\theta-\cos\theta=\dfrac{\sin^2\theta}{\cos\theta}\).
2. So \(x=\tan\theta\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(2\)
(b) \(1\)
(c) \(\frac12\)
(d) \(-1\)
Show Answer & Explanation
Answer: (a) \(2\)
Explanation:
1. \(\cot(A+B)=1\) gives \(\cot A\cot B-1=\cot A+\cot B\).
2. \((\cot A-1)(\cot B-1)=\cot A\cot B-\cot A-\cot B+1=2\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\frac1{25},\frac7{25},\frac17\)
(b) \(\frac{24}{25},\frac7{25},\frac{24}7\)
(c) \(\frac{24}{25},\frac1{25},\frac{24}{1}\)
(d) \(\frac1{25},\frac{24}{25},\frac1{24}\)
Show Answer & Explanation
Answer: (b) \(\frac{24}{25},\frac7{25},\frac{24}7\)
Explanation:
1. \(\cos A=\frac45\), so \(\sin2A=2\cdot\frac35\cdot\frac45=\frac{24}{25}\).
2. \(\cos2A=1-2\cdot\frac9{25}=\frac7{25}\) and \(\tan2A=\frac{24}7\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\frac1{11}\)
(b) \(\frac5{11}\)
(c) \( rac2{11}\)
(d) \(\frac3{11}\)
Show Answer & Explanation
Answer: (c) \( rac2{11}\)
Explanation:
1. \(\tan\alpha=2\). \(\cos\beta=-\frac35\), \(\sin\beta=\frac45\), \(\tan\beta=-\frac43\).
2. \(\tan(\alpha+\beta)=\dfrac{2-\frac43}{1+\frac83}=\dfrac{2/3}{11/3}=\dfrac2{11}\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\tan3A\tan2A\tan A\)
(b) \(-\tan3A\tan2A\tan A\)
(c) \(\tan A\tan2A-\tan2A\tan3A-\tan3A\tan A\)
(d) \(\tan3A+\tan2A+\tan A\)
Show Answer & Explanation
Answer: (a) \(\tan3A\tan2A\tan A\)
Explanation:
1. \(\tan3A=\tan(2A+A)=\dfrac{\tan2A+\tan A}{1-\tan2A\tan A}\).
2. Cross-multiplying gives \(\tan3A-\tan2A-\tan A=\tan3A\tan2A\tan A\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(1\)
(b) \(2\)
(c) \(4\)
(d) \(3\)
Show Answer & Explanation
Answer: (d) \(3\)
Explanation:
1. \(\tan2A=\dfrac{1}{1-\frac14}=\frac43\).
2. \(\tan(2A+B)=\dfrac{\frac43+\frac13}{1-\frac49}=\dfrac{5/3}{5/9}=3\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(a\)
(b) \(\frac ab\)
(c) \(b\)
(d) \(\frac{a^2}{b}\)
Show Answer & Explanation
Answer: (c) \(b\)
Explanation:
1. \(\cos2\theta=\dfrac{b^2-a^2}{a^2+b^2}\) and \(\sin2\theta=\dfrac{2ab}{a^2+b^2}\).
2. Sum \(=\dfrac{b(b^2-a^2)+2a^2b}{a^2+b^2}=b\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(\frac{2\pi}{36}\)
(b) \(\frac{5\pi}{36}\)
(c) \(\frac{3\pi}{36}\)
(d) \(\frac{4\pi}{36}\)
Show Answer & Explanation
Answer: (b) \(\frac{5\pi}{36}\)
Explanation:
1. \(25\times\frac\pi{180}=\frac{5\pi}{36}\).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
(a) \(-\frac45\) but not \(\frac45\)
(b) \(-\frac45\) or \(\frac45\)
(c) \(\frac45\) but not \(-\frac45\)
(d) \(\frac35\) or \(-\frac35\)
Show Answer & Explanation
Answer: (b) \(-\frac45\) or \(\frac45\)
Explanation:
1. \(\tan\theta\) is negative in quadrants II and IV; \(\sin\theta=\dfrac{\tan\theta}{\pm\sqrt{1+\tan^2\theta}}\) gives \(\pm\frac45\) (positive in II, negative in IV).
Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.
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Multiple Choice Questions (MCQs) for JEE Mathematics Trigonometric Functions
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FAQs
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