Question

Please answer Don't give irrelevant answer

Answer 1

Answer:

Step-by-step explanation:

3sinx+5cosx=5  

squaring on both sides

(3sinx+5cosx)2=52

9(sinx)2+25(cosx)2+30sinxcosx=25

let this equation be 1

now take the second equation as

5sinx−3cosx=a

squaring on both sides

25(sinx)2+9(cosx)2−30sinxcosx=a2

let this equation be 2

now add equation 1 and equation 2

we get

9(sinx)2+25(cosx)2+30sinxcosx+25(sinx)2+9(cosx)2−30sinxcosx=25+a2

take out 9 and 25 common and +30sinxcosx-30sinxcosx gets cancelled

9((sinx)2+(cosx)2)+25((sinx)2+(cosx)2)=25+a2

we know that (sinx)2+(cosx)2=1

9(1)+25(1)=25+a2

9+25=25+a2

34=25+a2

34−25=a2

a2=9

a=3

therefore      5sinx−3cosx=3

hope it helps you