draw a circle with an angle theta in the second quadrant and use it to prove that tan^2 theta+1 = sec^2 theta

Question

Tan^2 theta+1 = sec^ theta

Answer 1

Draw a triangle ABC with its base AB horizontal and angle CAB obtuse (bigger than 90 degrees). CAB represents theta. Extend AB to meet the perpendicular from C at D. CDA is a right-angled triangle. Tangent of CAD=CD/DA; secant of CAD=AC/DA=sqrt(CD^2+DA^2)/DA=sqrt((CD/DA)^2+1)=sqrt((tan(CAD))^2+1); so (sec(CAD))^2=(tan(CAD))^2+1.

CAD+CAB=180 deg. Tan(CAB)=tan(theta)=-tan(CAD) because tangent is negative in quadrant 2. But when we square we get (tan(CAB))^2=(tan(CAD))^2, so (sec(theta))^2=(tan(theta))^2+1.

 

Answer 2

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)