How do you prove 1+tan^2 (x) = sec^2 (x)?

May 2, 2016

Using the following:

• $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
• ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$
• $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

for $\cos \left(x\right) \ne 0$, we have

$1 + {\tan}^{2} \left(x\right) = {\cos}^{2} \frac{x}{\cos} ^ 2 \left(x\right) + {\left(\sin \frac{x}{\cos} \left(x\right)\right)}^{2}$

$= {\cos}^{2} \frac{x}{\cos} ^ 2 \left(x\right) + {\sin}^{2} \frac{x}{\cos} ^ 2 \left(x\right)$

$= \frac{{\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right)$

$= \frac{1}{\cos} ^ 2 \left(x\right)$

$= {\left(\frac{1}{\cos} \left(x\right)\right)}^{2}$

$= {\sec}^{2} \left(x\right)$
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