Question

If 3sina + 5cosa = 5 ; then prove that 3cosa-5sina =_+3

Answer 1

HELLO DEAR,

3sinA + 5cosA = 5

on squaring both side

we get,

9sin²A + 25cos²A + 30sinAcosA = 25

= 9(1 - cos²A) + 25(1 - sin²A) + 30sinAcosA = 25

= 9 - 9cos²A + 25 - 25sin²A + 30sinAcosA = 25

= 9 + 25 - 25 = 9cos²A + 25sin²A - 30sinAcosA

= (3cosA) + (5sinA) - 2×3×5sinAcosA = 9

= (3cosA - 5sinA)² = 9

= 3cosA - 5sinA = +-3

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer to the question :

Given us,

3sinA+5cosA=5
By squaring both side,
(3sinA+5cosA)²=5²

or, (3sinA)²+2.(3sinA).(5cosA)+(5cosA)²=25
{ (a+b)²=a²+2ab+b² }

or, 9sinA²+ 2.3.5.sinA.cosA+25cosA²=25

or, 9sinA²+30sinAcosA+25cosA²=25

or, 9(1-cosA²)+30sinAcosA+25(1-sinA²)=25 { sinA²=1-cosA² and cosA²=1-sinA² }

or, 9-9cosA²+30sinAcosA+25-25sinA²=25

or, 9-9cosA²+30sinAcosA+25-25sinA²-25=0

or, 9-9cosA²+30sinAcosA-25sinA²=0

or, -(9-9cosA²+30sinAcosA-25sinA²)=0

or, -9+9cosA²-30sinAcosA+25sinA²=0

or, 9cosA²-30sinAcosA+25sinA²=9

or, (3cosA)²-2.(3cosA).(5sinA²)+(5sinA)²=9
{ (a-b)²=a²-2ab+b² }

or, (3cosA-5sinA)²=9

or, 3cosA-5sinA= ±√9

or, 3cosA-5sinA=±3