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Fynjy0 [20]
3 years ago
6

Urgent!! Will mark brainliest!!

Mathematics
1 answer:
horsena [70]3 years ago
3 0

Answer:

1) x is negative and y is positive ⇒ last answer

2) cotФ = -12/35 ⇒ second answer

3) The right identity is cot²Ф - csc²Ф = -1 ⇒ last answer

Step-by-step explanation:

* For any point (x , y) lies on the terminal side of the angle Ф

 in standard position

* x = cosФ and y = sinФ

- If Ф in the first quadrant, then x , y are positive

∴ All trigonometry functions are positive

- If Ф in the second quadrant, then x is negative , y is positive

∴ sinФ only is positive

- If Ф in the third quadrant, then x is negative , y is negative

∴ tanФ only is positive

- If Ф in the fourth quadrant, then x is positive , y is negative

∴ cosФ only is positive

* Lets solve the problems

∵ Ф = 3π/4 ⇒ (135°)

∴ It lies on the second quadrant

∴ x is negative and y is positive

* Lets revise the reciprocal of sinФ, cosФ and tanФ

- cscФ = 1/sinФ

- secФ = 1/cosФ

- cotФ = 1/tanФ

∵ secФ = -37/12

∴ cosФ = -12/37

∵ π/2 < Ф < π

∴ Ф lies on the second quadrant

∴ cotФ is negative values

∵ tan²Ф = sec²Ф - 1

∵ secФ = -37/12

∴ tan²Ф = (-37/12)² - 1 = 1225/144 ⇒ take√ for both sides

∴ tanФ = ± 35/12

∵ cotФ = ± 12/35

∵ cotФ is negative value

∴ cotФ = -12/35

* In the standard position of the angle Ф the terminal

 of it lies on the unit circle O

- By using Pythagorean theorem

∵ x² + y² = 1

∵ x = cosФ and y = sinФ

∴ cos²Ф + sin²Ф = 1 ⇒ (1)

∴ cos²Ф = 1 - sin²Ф

∴ sin²Ф = 1 - cos²Ф

* Divide (1) by cos²Ф

∴ cos²Ф/cos²Ф + sin²Ф/cos²Ф = 1/cos²Ф

* Remember sin²Ф/cos²Ф = tan²Ф and 1/cos²Ф = sec²Ф

∴ 1 + tan²Ф = sec²Ф ⇒ (2) ⇒ subtract 1 from both sides

∴ tan²Ф = sec²Ф - 1 ⇒ subtract sec²Ф from both sides

∴ tan²Ф - sec²Ф = -1

* Divide (1) by sin²Ф

∴ cos²Ф/sin²Ф + sin²Ф/si²Ф = 1/sin²Ф

* Remember cos²Ф/sin²Ф = cot²Ф and 1/sin²Ф = csc²Ф

∴ cot²Ф + 1 = csc²Ф ⇒ (3) ⇒ subtract 1 from both sides

∴ cot²Ф = csc²Ф - 1 ⇒ subtract csc²Ф from both sides

∴ cot²Ф - csc²Ф = -1

* The right identity is cot²Ф - csc²Ф = -1

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