Question

Q. If tan A=2 , A lies in the 3rd quadrant
Find sin A + Cos A
ARE KOI TO BATAO.............!!!!
​

Answer 1

Answer:

I don't sis sry

Step-by-step explanation:

pls follow me

Answer 2

Answer:

Here given sinA=–4/5 and A lie in 3nd quadrant, i.e π<A<3π/2,lt means in 3nd quadrant cosA, sinA is negative and tanA is positive.

If sinA =–4/5

then,

cosA=-√(1-sin²A)=-√(1- 16/25) =-√(9/25) =-3/5

So,1-cosA=2sin²(A/2)

=>1- (-3/5)=2.sin²(A/2)

=>(1+ 3/5)=2sin²(A/2)

=>8/5=2sin²(A/2)

=>4/5 =sin²(A/2)

=>√(4/5)=sinA/2

=2/√5=sinA/2

cosec(A/2)=√5/2

Because π/2<A/2<3π/4 in 2nd quadrant.

Similarly,

sinA=2.sin(A/2).cos(A/2)

=>-4/5=2×2/√5 .cos(A/2)

-1/√5=cos(A/2)

Therefore,

tan(A/2) =sin(A/2)/cos(A/2)=2/√5 /(-1/√5) =-2

Step-by-step explanation:

Hope it's helpful to you ☺️ tanve