Question

If 41 sinA=40 show that tanA/tan^2A-1=360/1519

Answer 1

Given

`\sin(A) = (40)/(41)`

we can use the Pythagorean identity to find

`\cos(A) = \pm√(1 - \sin^2(A)) = \pm \frac9{41}`

Then

`\tan(A) = (\sin(A))/(\cos(A)) = \pm \frac{40}9`

`\tan^2(A) - 1 = \left(\pm \frac{40}9\right)^2 - 1 = (1519)/(81)`

`\implies (\tan(A))/(\tan^2(A) - 1) = \pm (360)/(1519)`

so the required equality holds as long as
`\cos(A)>0`.