If 3sinx + 4cosx = 5, then 4sinx - 3cosx =....???0-51/55plz explain me in detail... how to do questions like this...

ANS    3sinx+4cosx=5

     => 3/5 sinx +4/5 cosx = 1

     let cosA=3/5   => sinA=4/5

     =>  cosAsinx + sinAcosx = 1

     =>  sin(x+A) = 1

Now,

       4sinx - 3cosx

  =   5(4/5sinx - 3/5 cosx)     [multipying numerator and denominator by 5]

  =   5(sinAsinx - cosAcosx)

  =   -5{cos(x+A)} = -5[root{1-(sin(x+A)^2)}] = -5 x 0 = 0          Ans 0

TRICK: if asinx+bcosx is given then multiply numerator and denominator by root(a^2 +b^2)

            this method is useful  in many questions 

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Approved

3sinx+4cosx=5

Squaring on both sides.

(3sinx+4cosx) ² = 25

9sin²x +16 cos²x+24sinxcosx=25

9(1- cos²x)+16(1- sin²x)+ 24sinxcosx=25

9-9 cos²x+16-16 sin²x+24sinxcosx=25

25-9 cos²x-16 sin²x+24sinxcosx=25

9 cos²x+16 sin²x-24sinxcosx=0

(3cosx-4sinx) ² = 0

4sinx-3cosx=0