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.