Let
and
. Then
![m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab](https://tex.z-dn.net/?f=m%5E2-n%5E2%3D%28a%2Bb%29%5E2-%28a-b%29%5E2%3D%28a%5E2%2B2ab%2Bb%5E2%29-%28a%5E2-2ab%2Bb%5E2%29%3D4ab)
![\implies m^2-n^2=4\tan A\sin A](https://tex.z-dn.net/?f=%5Cimplies%20m%5E2-n%5E2%3D4%5Ctan%20A%5Csin%20A)
and
![mn=(a+b)(a-b)=a^2-b^2](https://tex.z-dn.net/?f=mn%3D%28a%2Bb%29%28a-b%29%3Da%5E2-b%5E2)
![\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}](https://tex.z-dn.net/?f=%5Cimplies4%5Csqrt%7Bmn%7D%3D4%5Csqrt%7B%5Ctan%5E2A-%5Csin%5E2A%7D)
The expression under the square root can be rewritten as
![\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)](https://tex.z-dn.net/?f=%5Ctan%5E2A-%5Csin%5E2A%3D%5Cdfrac%7B%5Csin%5E2A%7D%7B%5Ccos%5E2A%7D-%5Csin%5E2A%3D%5Csin%5E2A%5Cleft%28%5Cdfrac1%7B%5Ccos%5E2A%7D-1%5Cright%29%3D%5Csin%5E2A%28%5Csec%5E2A-1%29)
Recall that
![\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A](https://tex.z-dn.net/?f=%5Csin%5E2A%2B%5Ccos%5E2A%3D1%5Cimplies%5Ctan%5E2A%2B1%3D%5Csec%5E2A)
so that
![\tan^2A-\sin^2A=\sin^2A\tan^2A](https://tex.z-dn.net/?f=%5Ctan%5E2A-%5Csin%5E2A%3D%5Csin%5E2A%5Ctan%5E2A)
and assuming
and
, we end up with
![4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A](https://tex.z-dn.net/?f=4%5Csqrt%7B%5Ctan%5E2A-%5Csin%5E2A%7D%3D4%5Ctan%20A%5Csin%20A)
so that
![m^2-n^2=4\sqrt{mn}](https://tex.z-dn.net/?f=m%5E2-n%5E2%3D4%5Csqrt%7Bmn%7D)
as required.
Answer: X = -1 and Y = 3
—————————————
X + Y = 2
X = 2 - Y
————————
Y = x + 4
Y = 2 - Y + 4
2Y = 2 + 4
2Y = 6
Y = 3
————————
X + Y = 2
X + 3 = 2
X = -1
Answer:
if i labeled the triangle correctly BC= sqrt 97, ZX=32sqrt 2
Step-by-step explanation:
join my stream and tell me how to label the triangle
(ChildOfHermes)
X>44
.25(x)-3-4>4
.25(x)-7>4
.25(x)>11
x>11