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If 3sinA+ 4cosA=5 , then the value of 4sinA-3cosA=? Pls solve this fast

If 3sinA+ 4cosA=5 , then the value of 4sinA-3cosA=?
Pls solve this fast

Grade:11

5 Answers

Vikas TU
14149 Points
4 years ago
3sinA+ 4cosA=5
squaring both sides,
9sin2A + 16cos2B + 24sinAcosA = 25........................(1)
4sinA - 3cosA = x
squaring both sides,
16sin^2A + 9cos^2A – 24sinAcosA  = x^2 …............(2)
Add eqns. 1) and 2)
9 + 16  = x^2
x^2 = 25
x = +5  and  x = -5
raj
384 Points
4 years ago
hello,
3sinA+ 4cosA=5
squaring both sides,
9sin2A + 16cos2B + 24sinAcosA = 25........................(1)
4sinA - 3cosA = x
squaring both sides,
16sin^2A + 9cos^2A – 24sinAcosA  = x^2 …............(2)
Add eqns. 1) and 2)
9 + 16  = x^2
x^2 = 25
x = +5  and  x = -5
Kaunain
21 Points
4 years ago
Above ans is wrong..... Its... Ans should x =0.... By adding eq 1 and 2... 25 = 25 + x^20 = x^2So, x = 0.....ans....
hemu_gandi
11 Points
3 years ago
3 sinA + 4 cosA = 5Squaring on both sides. 9 sin² A+ 16 cos²A+24 sinA.cosA..........(1)Let,,4 sinA - 3 cos A = xThen,,16 sin² A+ 9 cos² A - 24 sinA.cosA=x²..........(2)Add (1)&(2)25(sin²A +cos² A)= 25+ x²25 = 25+x²X²=0x=0
Veniya sharma
32 Points
3 years ago
3sinx+4cosx=5-(1),4sinx-3cosx=y-(2)
Squaring and adding eq(1)&(2)
After solving,
25(sin^2x+cos^2x)=25+y^2
25=25+y^2
Y^2=0
Y=0

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