Given that $cos x = p$ $cos x =p$ find an expression, in terms of $p$ $p$ , for ${tan}^{2} x$ $tan^2 x$ ?

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Given that $cos x = p$ $cos x =p$ find an expression, in terms of $p$ $p$ , for ${tan}^{2} x$ $tan^2 x$ ?

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Answer 1 · 2015-10-12T14:38:17

${tan}^{2} x = \frac{1 - p^{2}}{p^{2}}$ $tan^2x= (1-p^2)/p^2$

Explanation:

If cos x =p, $sin x = \sqrt{1 - p^{2}}$ $sinx= sqrt(1-p^2)$

tan x = $\frac{\sqrt{1 - p^{2}}}{p}$ $sqrt(1-p^2)/p$

${tan}^{2} x = \frac{1 - p^{2}}{p^{2}}$ $tan^2x= (1-p^2)/p^2$

Answer 2 · 2015-10-12T21:35:47

${tan}^{2} x = \frac{1 - p^{2}}{p^{2}} = \frac{1}{p^{2}} - 1$ $tan^2 x = (1-p^2)/p^2 = 1/p^2 - 1$

Explanation:

If $cos x = p$ $cos x = p$ , then

${cos}^{2} x = p^{2}$ $cos^2 x = p^2$ .

By Pythagoras, ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2 x + cos^2 x = 1$ , so:

${sin}^{2} x = 1 - {cos}^{2} x = 1 - p^{2}$ $sin^2 x = 1 - cos^2 x = 1 - p^2$

Now $tan x = \frac{sin x}{cos x}$ $tan x = sin x / cos x$ , so

${tan}^{2} x = \frac{{sin}^{2} x}{{cos}^{2} x} = \frac{1 - p^{2}}{p^{2}} = \frac{1}{p^{2}} - 1$ $tan^2 x = sin^2 x / cos^2 x = (1 - p^2)/p^2 = 1/p^2 - 1$

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Given that $cos x = p$ $cos x =p$ find an expression, in terms of $p$ $p$ , for ${tan}^{2} x$ $tan^2 x$ ?

2 Answers

Explanation:

Explanation:

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