# Proof of trignometric identity

Could you help me prove this? I've gotten stuck, need some help..

$$\sin^2\Theta + \tan^2\Theta = \sec^2\Theta - \cos^2\Theta$$

Here's what I've done so far:

Left Side:

$$\sin^2\Theta + \frac{\sin^2\Theta}{\cos^2\Theta}=\frac{\sin^2\Theta\cos^2\Theta+\sin^2\Theta}{\cos^2\Theta}$$

Right Side:

$$\frac{1}{\cos^2\Theta} - \cos^2\Theta = \frac{1-\cos^2\Theta\cos^2\Theta}{\cos^2\Theta}$$

Note: I have tried simplifying it even further but I'm not getting the results, so I've left it at the points that I'm sure of.

• Consider moving the $\sin^2\Theta$ in the first line to the RHS and simplifying... – abiessu Dec 3 '15 at 4:03
• Try multiplying both side by $\cos^2\Theta$. Perhaps brush up on how to adding fractions as well. – John Joy Dec 3 '15 at 20:50
• Hint: $\sin^2\theta+\tan^2\theta = \sin^2\theta-1+1+\tan^2\theta$ – John Joy Dec 3 '15 at 20:52

$$\sin^2(\theta)+\tan^2(\theta) = \sec^2(\theta)-\cos^2(\theta) \\ \implies \left(\sin^2(\theta)+\cos^2(\theta)\right)+\tan^2(\theta) = \sec^2(\theta)$$ Can you proceed from here?

Hint: $\sin^2\theta+\cos^2\theta=1.$

Am I the only one here who was required to work on only one side when doing trig proofs?

The strategy for all such proofs is to work on both sides until you can get them to be equal, then retrace your steps and put them all on a single side.

To wit,

$$\sin^2x + \tan^2x = \sec^2x - \cos^2x$$
Add $\cos^2 x$ to both sides $$(\sin^2 x + \cos^2 x) + \tan^2 x = \sec^2 x$$

Using a trig identity gives
$$(1) + \tan^2 x = \sec^2 x$$

Which is itself an identity.

Now we achieved this after working on both sides, but this work reveals the "trick": let's use $\sin^2 x + \cos^2 x = 1$ on one side.

Working on the left hand side, we'll use the fundamental identity in this form: $\sin^2 x = 1 - \cos^2 x$.

$$\sin^2x + \tan^2x = \sec^2x - \cos^2x$$
$$(1 - \cos^2 x) + \tan^2 x = \sec^2x - \cos^2x$$
$$(1 + \tan^2 x) - \cos^2 x = \sec^2x - \cos^2x$$
$$\sec^2 x - \cos^2 x = \sec^2x - \cos^2x$$

Working on the right hand side is similar.