Question

if tanA=4 then find sinA cosA

Answer 1

Answer:

HELLO EVERYONE

YOUR ANSWER IS GIVEN BELOW :

given : tanA = 4

find : sinAcosA =?

now, we use identity -

1+tan²A=sec²A

1+(4) ²=sec²A

1+16=sec²A

17=sec²A

secA=√17

cosA=1/√17

now,

sin²A+cos²A=1

sin²A=1-cos²A

sin²A=1-(1/√17) ²

sin²A=1-1/17

sin²A=16/17

sinA=4/√17

sinA.cosA=4/√17 . 1/√17

sinA.cosA=4/17.

THANK YOU☺.

I HOPE IT'S HELPFUL FOR YOU.

Answer 2

tanA = 4 as given

We'll find sinAcosA

We know that tanθ = p/b

∴p/b = 4/1

Using Pythagoras Theorem,

h² = b² + p²

h² = 1 + 16

h = √17

sinA = p/h

sinA = 4/√17

cosA = b/h

cosA = 1/√17

∴sinAcosA = (4)/(√17) × (1)/(√17)

. = 4/17