Q. Solve
3SinA + 4CosA = 5
4CosA - 3SinA = ?

• Thank you for ur try sorry ur solution is wrong i become able to solve it
  i) 3cos(A) + 4sin(A) = 5 [Given]
  
  ii) Let 4CosA - 3SinA = x, the value of which needs to be found out.
  
  iii) Squaring the first one it is,
  9cos²A + 16sin²A + 24cos(A)sin(A) = 25
  4CosA - 3SinA = x
  Similarly squaring the second one,
  9sin²A + 16cos²A - 24cos(A)sin(A) = x²
  
  Adding both, 9 + 16 = 25 + x²
  [Since by identity, sin²A + cos²A = 1]
  
  ==> x² = 0
  So x = 0
  4CosA - 3SinA = 0
• 10 years agoHelpfull: Yes(3) No(3)
• Thank you for ur try sorry ur solution is wrong i become able to solve it
  i) 3cos(A) + 4sin(A) = 5 [Given]
  
  ii) Let 3sin(A) + 4cos(A) = x, the value of which needs to be found out.
  
  iii) Squaring the first one it is, 9cos²A + 16sin²A + 24cos(A)sin(A) = 25
  Similarly squaring the second one, 9sin²A + 16cos²A - 24cos(A)sin(A) = x²
  
  Adding both, 9 + 16 = 25 + x² [Since by identity, sin²A + cos²A = 1]
  
  ==> x² = 0
  So x = 0
  
  Thus 3sin(A) + 4cos(A) = 0
• 10 years agoHelpfull: Yes(2) No(4)
• @akanksha mishra your solution valid for other question
  u take
  i) 3cos(A) + 4sin(A) = 5 [Given]
  but in question 3SinA + 4CosA = 5
  solution for question
  
  3SinA + 4CosA = 5
  satisfy
  cosA=4/5,sinA=3/5
  so
  4CosA - 3SinA = =16/5-9/5=7/5
  ans is 7/5
  
  for your solution
  3cos(A) + 4sin(A) = 5
  cos a=3/5, sina 4/5
  4CosA - 3SinA = 4*3/5-3*4/5=0
  so please check your question
  
  
• 10 years agoHelpfull: Yes(1) No(0)
• 3SinA + 4CosA = 5
  or (3/5)sinA +(4/5)cosA = 1
  or cos(53)sinA + sin(53)cosA = 1 [since, cos(53 degree)=3/5]
  or sin(53+A)=1=sin(90)
  or 53+A=90
  or A=37
  so 4CosA - 3SinA
  = 4cos(37)-3sin(37)=4*(4/5)-3*(3/5)=7/5=1.4
• 10 years agoHelpfull: Yes(0) No(2)
• 3SinA + 4CosA = 5
  or (3/5)sinA +(4/5)cosA = 1
  let (3/5)=sinB then 4/5=cosB
  so eqn becomes
  sinB*sinA + cosB*cosA =1
  or cos(A-B)=1=cos(0 degree)
  or A-B=0
  or A=B
  SO (3/5)=sinB=sinA & (4/5)=cosB=cosA
  so 4CosA - 3SinA
  =4*(4/5)-3*(3/5)=7/5
  =1.4
  
• 10 years agoHelpfull: Yes(0) No(1)
• 3SinA + 4CosA = 5
  4CosA - 3SinA = x
  adding both gives, 8cosA=(5+x) or cosA=(5+x)/8
  subtraction gives, 6sinA=(5-x) or sinA=(5-x)/6
  now squaring & adding these two
  (cosA)^2+(sinA)^2=[(5+x)/8]^2+[(5-x)/6]^2
  or 1=[(5+x)/8]^2+[(5-x)/6]^2
  solving, x=1.4
  4CosA - 3SinA =x=1.4
• 10 years agoHelpfull: Yes(0) No(1)
• let x=sinA & Y=sinB
  then 3x+4y=5
  also x^2+y^2=1
  solving x=3/5,y=4/5
  so 4CosA - 3SinA =4y-3x=4*(4/5)-3*(3/5)=7/5=1.4
  
• 10 years agoHelpfull: Yes(0) No(1)
• 3 is opposite side of triangle and 4 is base and 5 is hypotenuse
  of right angled of triangle let be think.......
  Then....4cosA-3cosa= 7. =fix 1.2 answer
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  5
• 8 years agoHelpfull: Yes(0) No(0)