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.