draw a circle with an angle theta in the second quadrant and use it to prove that tan^2 theta+1 = sec^2 theta
To prove: tan^2(θ) + 1 = sec^2(θ)
From the diagram for the triangle OPQ, for the angle θ:
hyp (hypotenuse) = OP
adj (adjacent) = -OQ (distance is measured in negative x – direction)
opp (opposite) = PQ
tan(θ) = opp/adj = PQ/(-OQ)
tan(θ) = -(PQ/OQ)
sec(θ) = 1/cos(θ) = hyp/adj = OP/(-OQ)
sec(θ) = -(OP/OQ)
Using the above,
sec^2(θ) = OP^2/OQ^2
tan^2(θ) + 1 = PQ^2/OQ^2 + 1
tan^2(θ) + 1 = (PQ^2 + OQ^2)/OQ^2
By Pythagoras’ OP^2 = PQ^2 + OQ^2
Therefore tan^2(θ) + 1 = OP^2/OQ^2
I.e. tan^2(θ) + 1 = sec^2(θ)