Sum of Squares of Sine and Cosine

Theorem
$$\cos^2 x + \sin^2 x = 1$$

Corollaries

 * $$1 + \tan^2 x = \sec^2 x$$ (when $$\cos x \ne 1$$)


 * $$1 + \cot^2 x = \csc^2 x$$(when $$\sin x \ne 1$$)

Proof
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Corollaries

 * $$1 + \tan^2 x = \sec^2 x$$ (when $$\cos x \ne 1$$):

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 * $$1 + \cot^2 x = \csc^2 x$$(when $$\sin x \ne 1$$):

$$ $$ $$

Geometric Proof

 * Starting with $$\sin x$$ and $$\cos x$$,

$$\sin x = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$$

$$\cos x = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$$


 * Squaring both sides and adding them together gives:

$$\sin^2 x + \cos^2 x = \frac{\mathrm{opposite}^2 + \mathrm{adjacent}^2}{\mathrm{hypotenuse}^2} = 1$$ by Pythagoras's Theorem

Corollaries
Starting with $$\sin^2 x + \cos^2 x \equiv 1\!$$, divide every term by $$\cos^2 x \!$$ gives:

$$\tan^2 x + 1 = \sec^2 x \!$$

Alternatively, dividing by $$\sin^2 x \!$$ gives:

$$1 + \cot^2 x = \csc^2 x \!$$