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} ⇒ \frac{OM}{OP} = \frac{- 1}{2}.
Let OM = – 1 and OP = 2.
∴ 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} ⇒ \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} ⇒ \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 765°
7. cosec (- 1410°)
8. tan \frac{19π}{3}
9. sin (\frac{-11π}{3})
10. cot (\frac{-15π}{4})
Solutions to questions 6 – 10:
6. sin 765° = (8 × 90° + 45°)
= sin 45° = \frac{1}{\sqrt{2}}.
7. cosec (- 1410°) = – cosec 1410° [∵ cosec (-θ) = – cosec θ
= – cosec (16 × 90° – 30°)
= – cosec (- 30)° = – [- cosec 30°]
= cosec 30° = 2.
8. tan \frac{19π}{3} = tan (6π + \frac{π}{3}) = tan \frac{π}{3} = \sqrt{3}.
9. sin (\frac{- 11π}{3}) = – sin \frac{11π}{3} [∵ sin(-θ) = – sin θ]
= – sin(4π – \frac{π}{3}) = – (- sin \frac{π}{3})
= sin \frac{π}{3} = \frac{\sqrt{3}}{2}.
10. cot(\frac{- 15π}{4}) = – cot \frac{15π}{4} [∵ cot(-θ) = – cot θ]
= – cot (4π – \frac{π}{4}) = – cot (- \frac{π}{4})
= – (- cot \frac{π}{4}) = cot \frac{π}{4} = 1.