Question #7c148

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Question #7c148

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Answer 1 · 2018-02-22T03:58:36

$\Rightarrow \frac{1}{{cos}^{2} x} = {sec}^{2} x = 1 + {tan}^{2} x$ $color(green)(=> 1/ cos ^2 x = sec ^2 x = 1 + tan ^2 x$

Explanation:

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$tan x = \frac{sin x}{cos x}, csc x = \frac{1}{sin x}, \frac{1}{cos x} = sec x$ $tan x = sin x / cos x, csc x = 1/sin x, 1/cos x = sec x$

${tan}^{2} x \cdot {csc}^{2} x = (\frac{{sin}^{2} x}{{cos}^{2} x}) \cdot (\frac{1}{{sin}^{2} x})$ $tan^2x * csc^2 x = (color(brown)cancel(sin^2 x) / (cos^2 x)) * (1/ (color(brown)cancel(sin^2 x)))$

$=> 1/ cos ^2 x = sec ^2 x = 1 + tan ^2 x$

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Question #7c148

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