Question #f5744

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Question #f5744

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Answer 1 · 2017-02-16T19:47:45

With questions like these we usually exploit the difference of squares identity:

$x^{2} - y^{2} = (x + y) (x - y)$ $x^2-y^2=(x+y)(x-y)$

Multiply numerator and denominator of $L H S$ $LHS$ of the trig expression by $(1 + sin u)$ $(1+sinu)$

Explanation:

$\frac{1 + sin u}{1 - sin u} \times \frac{1 + sin u}{1 + sin u}$ $(1+sinu)/(1-sinu)xx(1+sinu)/(1 +sinu)$

$sin u \neq 1$ $sinu!=1$

$= \frac{1 + 2 sin u + {sin}^{2} u}{1 - {sin}^{2} u}$ $=(1+2sinu+sin^2u)/(1-sin^2u)$

$but 1 - {sin}^{2} u = {cos}^{2} u$ $"but " 1-sin^2u=cos^2u$

$= \frac{1 + 2 sin u + {sin}^{2} u}{{cos}^{2} u}$ $=(1+2sinu+sin^2u)/cos^2u$

$= \frac{1}{{cos}^{2} u} + 2 \frac{sin u}{{cos}^{2} u} + \frac{{sin}^{2} u}{{cos}^{2} u}$ $=1/cos^2u+2sinu/cos^2u+sin^2u/cos^2u$

$= \frac{1}{{cos}^{2} u} + 2 \frac{sin u}{cos u} \times \frac{1}{cos u} + \frac{{sin}^{2} u}{{cos}^{2} u}$ $=1/cos^2u+2color(red)(sinu/cosu)xxcolor(blue)(1/cosu)+sin^2u/cos^2u$

$= {sec}^{2} u + 2 tan u sec u + {tan}^{2} u$ $=sec^2u+2color(red)(tanu)color(blue)(secu)+tan^2u$

$= {(sec u + tan u)}^{2} as required$ $=(secu+tanu)^2" as required"$

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Question #f5744

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