Assuming that by sin^(-1)(-7/25)sin−1(−725) you mean arcsin(-7/25)arcsin(−725),
We know the Pythagorean identity sin^2(theta) + cos^2(theta) =1sin2(θ)+cos2(θ)=1, so by dividing both sides by sin^2(theta)sin2(θ) we have
sin^2(theta)/sin^2(theta) + cos^2(theta)/sin^2(theta) = 1/sin^2(theta) rarr 1 + 1/tan^2(theta) = 1/sin^2(theta)sin2(θ)sin2(θ)+cos2(θ)sin2(θ)=1sin2(θ)→1+1tan2(θ)=1sin2(θ)
Substitute thetaθ for arcsin(-7/25)arcsin(−725)
1 + 1/tan^2(arcsin(-7/25)) = 1/sin^2(arcsin(-7/25))1+1tan2(arcsin(−725))=1sin2(arcsin(−725))
Knowing that sin(arcsin(theta)) = thetasin(arcsin(θ))=θ we have
1 + 1/tan^2(arcsin(-7/25)) = 1/(-7/25)^2 = 625/491+1tan2(arcsin(−725))=1(−725)2=62549
Isolating the tangent
1/tan^2(arcsin(-7/25)) = 625/49 - 11tan2(arcsin(−725))=62549−1
Take the least common factor to sum the two fractions
1/tan^2(arcsin(-7/25)) = (625-49)/49 = 576/491tan2(arcsin(−725))=625−4949=57649
Invert both sides
tan^2(arcsin(-7/25)) = 49/576tan2(arcsin(−725))=49576
Take the square root
tan(arcsin(-7/25)) = +-sqrt(49/576)tan(arcsin(−725))=±√49576
Simplifying / Evaluating
tan(arcsin(-7/25)) = +-7/sqrt(2*2*12*12) = +-7/24tan(arcsin(−725))=±7√2⋅2⋅12⋅12=±724
We know from the domain of the arcsin that the angles with negative sines are on the fourth quadrant, so it's negative.
tan(arcsin(-7/25)) = -7/24tan(arcsin(−725))=−724
Or, use the main Pythagorean identity
Knowing that sin(arcsin(theta)) = thetasin(arcsin(θ))=θ, we have that sin(arcsin(-7/25)) = -7/25sin(arcsin(−725))=−725
We know that cos^2(theta) = 1 - sin^2(theta)cos2(θ)=1−sin2(θ), so
cos^2(arcsin(-7/25)) = 1 - 49/625 = 576/625cos2(arcsin(−725))=1−49625=576625
cos(arcsin(-7/25)) = +-24/25cos(arcsin(−725))=±2425
So tan(theta) = sin(theta)/cos(theta)tan(θ)=sin(θ)cos(θ) or
tan(arcsin(-7/25)) = -7/25*(+-25/24) = +-(-7/24)tan(arcsin(−725))=−725⋅(±2524)=±(−724)
We know from the domain of the arcsin that the angles with negative sines are on the fourth quadrant, so the tangent is negative.
tan(arcsin(-7/25)) = -7/24tan(arcsin(−725))=−724
The first way is good to get a feel of how to mess with the less common Pythagorean identities that could be needed in harder problems whereas the second is good for quick plug and chugs kind of problems.
