 # Trig Identities

 From: Nicolas Lawton Date: 23 August 1999 Subject: Prove this trig identity Can you help me prove the identity: tan2A secA = 2 sec2A sinA

### Maths Help suggests:

Proving trigonometric identites like this is a common A-level Pure Maths exercise. You should always have your booklet of standard trig formulae in front of you to remind you of what can be used.

Basically, you start with the left hand side and rearrange it step by step until you get the right hand side. So let's look at the left hand side:
tan2A secA
You might spot the double angle and think you should replace tan2A by (2tanA)/(1-tan2A). Normally this would be a good starting point. BUT take a sneak peek at what you are aiming for on right hand side. That still contains the double angle 2A. So it would NOT be a good idea here to replace the tangent of a double angle by the tangents of single angles.

Instead, use the fact that tanX = sinX/cosX. This makes the left hand side into:
(sin2A / cos2A) secA Now this contains a fraction. We don't want a fraction in the answer. Remember that the reciprocal of cosX can be written as secX. We therefore get:
sin2A sec2A secA

Aha! The "sec2A" part is fine, because that appears on the right hand side. So now focus on the remaining "sin2A secA". Now it IS appropriate to replace the double angle. Use the fact that sin2A = 2sinAcosA and we get:
2 sinA cosA sec2A secA

Take another peek at what we are aiming for on the right hand side. Well, it is all there, if we can just clear away the cosA and secA. But again, remember that secA is just another name for 1/cosA. Therefore cosA*secA = 1 meaning that the cosA and secA "cancel out". So we get
2 sinA sec2A
which is the right hand side as required. DONE!!

When considering identities, it is a good idea to use your graphics calculator to plot the graphs of the LHS and the RHS. They should (obviously!) be indistinguishable - one should lie exactly on top of the other. Then every time you re-arrange the LHS, plot the graph of what you get. If you have done it right, the graph should again be indistinguishable. If the graph looks different, check your work!!

On axes x=0 to 360 degrees and y=-5 to 5 the graph should like like that below. (Note the asymptotes at x=45, 135, 225, 315 degrees.) Return to Trigonometry & Geometry contents list