Question

iii) If 3 sin a +5 cos a = 5 then show that
5 sin à - 3 cos a = 3​

Answer 1

3sinA+5cosA=5

squaring on both sides, we get

(3sinA+5cosA)²=(5)²

9sin²A+30sinAcosA+25cos²A=25 [(a+b)²=a²+2ab+b²]

9(1-cos²A)+30sinAcosA+25(1-sin²A)=25

9-9cos²A+30sinAcosA+25-25sin²A=25

34-9cos²A+30sinAcosA-25sin²A=25

-9cos²A+30sinAcosA-25sin²A=25-34

-9cos²A+30sinAcosA-25sin²A= -9

changing the sign.

9cos²A-30sinAcosA+25sin²A=9

we can write,

25sin²A-30sinAcosA+9cos²A=9 [a²-2ab+b²=(a-b)]

(5sinA-3cosA)²=9

5sinA-3cosA=√9

5sinA-3cosA=3

H.P