Question

Prove that, 2sin^2A+5cosA=4

Answer 1

2sin^2A + 5 cosA=4
2(sin^2A)+5 cosA=4
2(1-cos^2A)+5 cosA=4
[sin^A= 1-cos^A]
2-2cos^2A+5 cosA=4 .......(i)
-2cos^2A+5 cosA=4-2
-2cos^2A+5 cosA-2=0
-1(2cos^2A-5 cosA+2)=0
2cos^2A-5 cosA+2=0
[by factorization],
2cos^2A - 4cosA - cosA +2=0
2cosA(cosA - 2) -1 (cosA - 2)=0
(2cosA - 1)=0 & (cosA - 2)=0
2cosA=1 & cosA=2
cosA=1/2 & cosA=2

then put the values of cosA in the equation(i) we get,

2-2cos^2A+5 cosA=4
2-2(1/2)^2+5 (1/2)=4
2-2(1/4)+(5/2)=4
2-(1/2)+(5/2)=4
$\frac{2}{1} - \frac{1}{2} + \frac{5}{2} = 4$
$\frac{8}{2} = 4$
so,
4=4
proved.

if this help then plz mark as brainliest...........:)