Exercise 3.2 — Trigonometric Functions

1. 3.2.13 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   cos x = -1/2, x lies in the third quadrant. Find the values of the other five trigonometric functions.

   Hint. Get sin x from sin squared x + cos squared x = 1, then fix its sign using the third quadrant before finding the rest.

   sin^2 x = 1 - 1/4 = 3/4, so sin x = +-root3/2. In the third quadrant sin x is negative, so sin x = -root3/2. Then tan x = sinx/cosx = root3, cot x = 1/root3, sec x = 1/cosx = -2, cosec x = 1/sinx = -2/root3.

   ✦ Working through each part gives: sin x = -root3/2, tan x = root3, cot x = 1/root3, sec x = -2, cosec x = -2/root3.

2. 3.2.23 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   sin x = 3/5, x lies in the second quadrant. Find the values of the other five trigonometric functions.

   Hint. Get cos x from the Pythagorean identity, then fix its sign using the second quadrant.

   cos^2 x = 1 - 9/25 = 16/25, so cos x = +-4/5. In the second quadrant cos x is negative, so cos x = -4/5. Then tan x = (3/5)/(-4/5) = -3/4, cot x = -4/3, sec x = -5/4, cosec x = 5/3.

   ✦ Working through each part gives: cos x = -4/5, tan x = -3/4, cot x = -4/3, sec x = -5/4, cosec x = 5/3.

3. 3.2.33 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   cot x = 4/3, x lies in the third quadrant. Find the values of the other five trigonometric functions.

   Hint. tan x = 1/cot x gives the magnitude of the ratio; a right triangle with legs 3 and 4 gives the reference values, and the third quadrant fixes both sin x and cos x as negative.

   tan x = 1/cotx = 3/4 (positive, consistent with the third quadrant). Using a 3-4-5 reference triangle: sin x = -3/5 and cos x = -4/5 (both negative in QIII). Then sec x = 1/cosx = -5/4, cosec x = 1/sinx = -5/3.

   ✦ Working through each part gives: sin x = -3/5, cos x = -4/5, tan x = 3/4, sec x = -5/4, cosec x = -5/3.

4. 3.2.43 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   sec x = 13/5, x lies in the fourth quadrant. Find the values of the other five trigonometric functions.

   Hint. cos x = 1/sec x immediately; get sin x from the Pythagorean identity and fix its sign using the fourth quadrant.

   cos x = 5/13. sin^2 x = 1 - 25/169 = 144/169, so sin x = +-12/13. In the fourth quadrant sin x is negative, so sin x = -12/13. Then tan x = -12/5, cot x = -5/12, cosec x = -13/12.

   ✦ Working through each part gives: cos x = 5/13, sin x = -12/13, tan x = -12/5, cot x = -5/12, cosec x = -13/12.

5. 3.2.53 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   tan x = -5/12, x lies in the second quadrant. Find the values of the other five trigonometric functions.

   Hint. A 5-12-13 reference triangle gives the magnitudes; the second quadrant fixes sin x positive and cos x negative.

   Using a 5-12-13 reference triangle with the second quadrant: sin x = 5/13 (positive), cos x = -12/13 (negative). Then cot x = -12/5, sec x = 1/cosx = -13/12, cosec x = 1/sinx = 13/5.

   ✦ Working through each part gives: sin x = 5/13, cos x = -12/13, cot x = -12/5, sec x = -13/12, cosec x = 13/5.

6. 3.2.62 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Find the value of sin 765°.

   Hint. Strip off whole revolutions (multiples of 360 degrees) first.

   765deg = 2 x 360deg + 45deg = 720deg + 45deg. By periodicity, sin765deg = sin45deg = 1/root2.

   ✦ Working through each part gives: 1/root2.

7. 3.2.72 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Find the value of cosec(-1410°).

   Hint. Add a whole multiple of 360 degrees to bring the angle into a familiar range, then evaluate.

   -1410deg + 4 x 360deg = -1410deg + 1440deg = 30deg. By periodicity, cosec(-1410deg) = cosec(30deg) = 1/sin30deg = 2.

   ✦ Working through each part gives: 2.

8. 3.2.82 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Find the value of tan(19π/3).

   Hint. Subtract a whole multiple of 2 pi (a multiple of 6pi/3) to bring the angle between 0 and 2pi.

   19pi/3 - 6pi = 19pi/3 - 18pi/3 = pi/3. By periodicity, tan(19pi/3) = tan(pi/3) = root3.

   ✦ Working through each part gives: root3.

9. 3.2.92 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Find the value of sin(-11π/3).

   Hint. Add a whole multiple of 2 pi directly to the angle -11pi/3 itself before evaluating -- periodicity applies to the angle as given, sign included.

   -11pi/3 + 4pi = -11pi/3 + 12pi/3 = pi/3. By periodicity, sin(-11pi/3) = sin(pi/3) = root3/2.

   ✦ Working through each part gives: root3/2.

   Where students slip. Separating the negative sign first and using sin(-theta) = -sin(theta) before reducing the angle often leads to a sign slip -- it is simpler and safer to add the multiple of 2pi to the angle exactly as given.

10. 3.2.102 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

    Find the value of cot(-15π/4).

    Hint. Add a whole multiple of 2 pi (i.e. a multiple of 8pi/4) to bring the angle into a familiar range.

    -15pi/4 + 4pi = -15pi/4 + 16pi/4 = pi/4. By periodicity, cot(-15pi/4) = cot(pi/4) = 1.

    ✦ Working through each part gives: 1.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh103.pdf) — three numbered exercises (3.1-3.3, 42 questions) plus the chapter's Miscellaneous Exercise (10 questions); the chapter has no Exercise 3.4 and no general-solution-of-trigonometric-equations content, contrary to an earlier stub. Miscellaneous Exercise Q1's stacked-fraction identity was re-rendered directly from the PDF at 300dpi after raw text extraction garbled the numerators, confirming it as 2cos(pi/13)cos(9pi/13)+cos(3pi/13)+cos(5pi/13)=0. Questions are referenced from the NCERT textbook for identification.