Trigonometry

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Trigonometry — solved practice questions

8 JEE Mathematics questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   The maximum value of 3 cos x + 4 sin x + 7 is:

   • A7
   • B10
   • C12
   • D14
   Show answer & solution

   Correct answer: (C) 12

   For a cos x + b sin x the maximum is sqrt(a^2 + b^2). Here sqrt(9 + 16) = 5, so the maximum of the whole expression is 5 + 7 = 12.

2. Q2easy

   The value of cos 15 degrees is:

   • A(sqrt6 + sqrt2)/4
   • B(sqrt6 - sqrt2)/4
   • C(sqrt3 + 1)/2
   • D(sqrt3 - 1)/(2 sqrt2)
   Show answer & solution

   Correct answer: (A) (sqrt6 + sqrt2)/4

   cos 15 = cos(45 - 30) = cos45 cos30 + sin45 sin30 = (sqrt6 + sqrt2)/4.

3. Q3easy

   If sin A = 3/5 and A is acute, then cos 2A equals:

   • A24/25
   • B-7/25
   • C18/25
   • D7/25
   Show answer & solution

   Correct answer: (D) 7/25

   cos 2A = 1 - 2 sin^2 A = 1 - 2(9/25) = 1 - 18/25 = 7/25.

4. Q4medium

   If tan A = 1/2 and tan B = 1/3, where A and B are acute, then A + B equals:

   • Api/3
   • Bpi/4
   • Cpi/2
   • Dpi/6
   Show answer & solution

   Correct answer: (B) pi/4

   tan(A + B) = (1/2 + 1/3)/(1 - 1/6) = (5/6)/(5/6) = 1, so A + B = pi/4.

5. Q5medium

   If sec x - tan x = 4, then sec x + tan x equals:

   • A1/4
   • B4
   • C-1/4
   • D1/2
   Show answer & solution

   Correct answer: (A) 1/4

   Since sec^2 x - tan^2 x = 1, we have (sec x - tan x)(sec x + tan x) = 1, so sec x + tan x = 1/4.

6. Q6medium

   If sin x + cos x = 1/5 and x lies in the second quadrant, then tan x equals:

   • A-3/4
   • B3/4
   • C-4/3
   • D4/3
   Show answer & solution

   Correct answer: (C) -4/3

   Squaring gives 1 + 2 sin x cos x = 1/25, so sin x cos x = -12/25. Then (sin x - cos x)^2 = 49/25 giving sin x - cos x = 7/5 (positive in Q2), so sin x = 4/5, cos x = -3/5 and tan x = -4/3.

7. Q7medium

   In a triangle with sides a = 7, b = 8, c = 9, the value of cos A is:

   • A1/2
   • B2/3
   • C3/4
   • D11/21
   Show answer & solution

   Correct answer: (B) 2/3

   By the cosine rule cos A = (b^2 + c^2 - a^2)/(2bc) = (64 + 81 - 49)/(2·8·9) = 96/144 = 2/3.

8. Q8medium

   The general solution of the equation 2 cos^2 x + 3 sin x = 0 is (n is any integer):

   • An·pi + (-1)^n (pi/6)
   • B2n·pi ± (pi/3)
   • Cn·pi ± (pi/6)
   • Dn·pi + (-1)^n (-pi/6)
   Show answer & solution

   Correct answer: (D) n·pi + (-1)^n (-pi/6)

   Using cos^2 x = 1 - sin^2 x gives 2 sin^2 x - 3 sin x - 2 = 0, so sin x = 2 (rejected) or sin x = -1/2. Hence x = n·pi + (-1)^n (-pi/6).

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