# GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2

Find the values of the other five trigonometric functions in questions 1 to 5:
1. cos x = – $$\frac{1}{2}$$, x lies in third quadrant.
2. sin x = $$\frac{3}{5}$$, x lies in second quadrant.
3. cot x = $$\frac{3}{4}$$, x lies in third quadrant.
4. sec x = $$\frac{13}{5}$$, x lies in fourth quardrant.
5. tan x = – $$\frac{5}{12}$$, x lies in the second quadrant.
Solutions to questions 1 – 5:
1. Since x lies in the 3rd quadrant, therefore
cos x = – $$\frac{1}{2}$$ â‡’ $$\frac{OM}{OP}$$ = $$\frac{- 1}{2}$$.
Let OM = – 1 and OP = 2.
âˆ´ MP = $$\sqrt{\mathrm{OP}^{2}-\mathrm{OM}^{2}}$$ = – $$\sqrt{4 – 1}$$ = – $$\sqrt{3}$$

2. Since x lies in the second quadrant, therefore
sin x = $$\frac{3}{5}$$ â‡’ $$\frac{MP}{OM}$$ = $$\frac{3}{5}$$
Let MP = 3, S0, OP = 5.

3. Since x lies in third quadrant, therefore
cot x = $$\frac{3}{4}$$ = $$\frac{- 3}{- 4}$$
Let MP = – 4. So, OM = – 3. Then,

4. Since x lies in fourth quardrant, therefore
sec x = $$\frac{13}{5}$$ â‡’ $$\frac{OP}{OM}$$ = $$\frac{13}{5}$$ [Given]
Let OP = 13. So, OM = 5. Then,

5. x lies in the second quadrant

Questions?
Find the values of the following trigonometric ratios:
6. sin 765Â°
7. cosec (- 1410Â°)
8. tan $$\frac{19Ï€}{3}$$
9. sin ($$\frac{-11Ï€}{3}$$)
10. cot ($$\frac{-15Ï€}{4}$$)
Solutions to questions 6 – 10:
6. sin 765Â° = (8 Ã— 90Â° + 45Â°)
= sin 45Â° = $$\frac{1}{\sqrt{2}}$$.

7. cosec (- 1410Â°) = – cosec 1410Â° [âˆµ cosec (-Î¸) = – cosec Î¸
= – cosec (16 Ã— 90Â° – 30Â°)
= – cosec (- 30)Â° = – [- cosec 30Â°]
= cosec 30Â° = 2.

8. tan $$\frac{19Ï€}{3}$$ = tan (6Ï€ + $$\frac{Ï€}{3}$$) = tan $$\frac{Ï€}{3}$$ = $$\sqrt{3}$$.

9. sin ($$\frac{- 11Ï€}{3}$$) = – sin $$\frac{11Ï€}{3}$$ [âˆµ sin(-Î¸) = – sin Î¸]
= – sin(4Ï€ – $$\frac{Ï€}{3}$$) = – (- sin $$\frac{Ï€}{3}$$)
= sin $$\frac{Ï€}{3}$$ = $$\frac{\sqrt{3}}{2}$$.

10. cot($$\frac{- 15Ï€}{4}$$) = – cot $$\frac{15Ï€}{4}$$ [âˆµ cot(-Î¸) = – cot Î¸]
= – cot (4Ï€ – $$\frac{Ï€}{4}$$) = – cot (- $$\frac{Ï€}{4}$$)
= – (- cot $$\frac{Ï€}{4}$$) = cot $$\frac{Ï€}{4}$$ = 1.