Question #cef51

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Question #cef51

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Answer 1 · 2017-08-19T13:09:33

See the solution below

Explanation:

L.H.S
$= {tan}^{2} (x) + {cos}^{2} (x) \cdot \frac{1}{sec (x) + sin (x)}$ $= tan^2(x) + cos^2(x) * 1/ (sec(x) + sin(x))$
$= {sec}^{2} (x) - 1 + {cos}^{2} (x) \cdot \frac{1}{sec (x) + sin (x)}$ $= sec^2(x) -1 + cos^2(x) * 1/ (sec(x) + sin(x))$

[since ${tan}^{2} (x) = {sec}^{2} (x) - 1$ $color(blue)(tan^2(x) = sec^2(x)-1$ )]

$= {sec}^{2} (x) - 1 + 1 - {sin}^{2} (x) \cdot \frac{1}{sec (x) + sin (x)}$ $= sec^2(x) -1 +1 - sin^2(x) * 1/(sec(x) + sin(x))$
$= {sec}^{2} (x) - {sin}^{2} (x) \cdot \frac{1}{sec (x) + sin (x)}$ $= sec^2(x) - sin^2(x) * 1/(sec(x) + sin(x))$
$= {sec (x) + sin (x)} {sec (x) - sin (x)} \cdot \frac{1}{sec (x) + sin (x)}$ $= {sec(x) + sin(x)}{sec(x) - sin(x)}*1/(sec(x) + sin(x))$

simplifying we get as
$= sec (x) - sin (x) = R . H . S$ $= sec(x) - sin(x) = R.H.S$

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Question #cef51

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