How do you verify $\frac{1 + {sec}^{2} x}{1 + {tan}^{2} x} = 1 + {cos}^{2} x$ $(1 + sec^2 x) / (1 + tan^2 x) = 1 + cos^2 x$ ?

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How do you verify $\frac{1 + {sec}^{2} x}{1 + {tan}^{2} x} = 1 + {cos}^{2} x$ $(1 + sec^2 x) / (1 + tan^2 x) = 1 + cos^2 x$ ?

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Answer 1 · 2016-07-03T16:21:47

Use the Pythagorean Identity $1 + {tan}^{2} x = {sec}^{2} x$ $1+tan^2x=sec^2x$ .

Explanation:

Recall the Pythagorean Identity $1 + {tan}^{2} x = {sec}^{2} x$ $1+tan^2x=sec^2x$ (this can be derived by dividing the identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ by ${cos}^{2} x$ $cos^2x$ ). The key to this problem is applying this identity.

Since $1 + {tan}^{2} x = {sec}^{2} x$ $1+tan^2x=sec^2x$ , we can replace the $1 + {tan}^{2} x$ $1+tan^2x$ in the denominator with ${sec}^{2} x$ $sec^2x$ :
$\frac{1 + {sec}^{2} x}{{sec}^{2} x} = 1 + {cos}^{2} x$ $(1+sec^2x)/(sec^2x)=1+cos^2x$

Now we can break the fraction up in two:
$\frac{1}{{sec}^{2} x} + \frac{{sec}^{2} x}{{sec}^{2} x} = 1 + {cos}^{2} x$ $1/sec^2x+sec^2x/sec^2x=1+cos^2x$

Since $\frac{1}{sec x} = cos x$ $1/secx=cosx$ , $\frac{1}{{sec}^{2} x} = {cos}^{2} x$ $1/sec^2x=cos^2x$ ; and $\frac{{sec}^{2} x}{{sec}^{2} x} = 1$ $sec^2x/sec^2x=1$ . So:
${cos}^{2} x + 1 = 1 + {cos}^{2} x$ $cos^2x+1=1+cos^2x$

Using the commutative property of addition we can rearrange the left side of the equation to match the right:
$1 + {cos}^{2} x = 1 + {cos}^{2} x$ $1+cos^2x=1+cos^2x$

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How do you verify $\frac{1 + {sec}^{2} x}{1 + {tan}^{2} x} = 1 + {cos}^{2} x$ $(1 + sec^2 x) / (1 + tan^2 x) = 1 + cos^2 x$ ?

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How do you verify $\frac{1 + {sec}^{2} x}{1 + {tan}^{2} x} = 1 + {cos}^{2} x$ $(1 + sec^2 x) / (1 + tan^2 x) = 1 + cos^2 x$ ?