Question

sina+cosa=0 if a is in 4th quadrant then find sina and cosa​

Answer 1

sinA + cosA=0

⇒sinA = -cosA

Dividing cosA to both sides :

⇒(sinA)/(cosA) = -1

⇒tanA = -1                                     ...(i)

Now,

We know, sec²A = 1 + tan²A

⇒sec²A = 1 + (-1)²                         [from (i)]

⇒sec²A = 1 + 1

⇒secA = |√2|

As sec is positive in the 4th quadrant, so the magnitude becomes positive.

secA = √2

So, cosA = 1/secA

⇒cosA = 1/√2

We know,

tanA * cosA = (sinA/cosA) * cosA = sinA

sinA = -1*1/√2

⇒sinA = -1/√2

Answer 2

ANSWER:

Given

• sinA + cosA = 0
• A is in 4th quadrant
• sinA = ? , cosA = ?

➔ sinA + cosA = 0

Simplifying the given equation

➔ sinA = - cosA

➔ sinA = - 1(cosA)

➔ sinA/cosA = -1

➔ tanA = -1

Using

tanA = ±√sec²A - 1

➔ √sec²A - 1 = - 1

➔ sec²A - 1 = (-1)²

➔ sec²A = 1 + 1

➔ sec²A = 2

➔ secA = √2

➔ 1/cosA = √2

➔ cosA = 1/√2

Given A is in 4 th quadrant then cos(-A) is +ve

➔ cosA = 1/√2

Using

cosA = ±√1 - sin²A

➔ √1 - sin²A = -1/√2

➔ 1 - sin²A = (- 1/√2)²

➔ 1 - sin²A = 1/2

➔ 1 - 1/2 = sin²A

➔ 1/2 = sin²A

➔ √1/2 = sinA

➔ sinA = 1/√2

Given A is in 4th quadrant then sinA is -ve

➔ sinA = -1/√2

A = - 45°

Hence, sinA = -1/√2 , cosA = 1/√2