Prove that $\frac{1 + sin x}{1 - sin x} = {(sec x + tan x)}^{2}$ $(1+sinx)/(1-sinx)=(secx+tanx)^2$ ?

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Prove that $\frac{1 + sin x}{1 - sin x} = {(sec x + tan x)}^{2}$ $(1+sinx)/(1-sinx)=(secx+tanx)^2$ ?

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Answer 1 · 2016-10-13T10:27:28

Using the identities

${sin}^{2} (x) + {cos}^{2} (x) = 1 \Rightarrow 1 - {sin}^{2} (x) = {cos}^{2} (x)$ $sin^2(x)+cos^2(x) = 1 => 1-sin^2(x) = cos^2(x)$
$(a - b) (a + b) = a^{2} - b^{2}$ $(a-b)(a+b) = a^2-b^2$

we have

$\frac{1 + sin (x)}{1 - sin (x)} = \frac{(1 + sin (x)) (1 + sin (x))}{(1 - sin (x)) (1 + sin (x))}$ $(1+sin(x))/(1-sin(x)) = ((1+sin(x))(1+sin(x)))/((1-sin(x))(1+sin(x)))$

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

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

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

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

$= {(sec (x) + tan (x))}^{2}$ $=(sec(x)+tan(x))^2$

Answer 2 · 2016-10-13T10:27:46

$\frac{1 + sin x}{1 - sin x} = {(sec x + tan x)}^{2}$ $(1+sinx)/(1-sinx)=(secx+tanx)^2$

Explanation:

Let us start from right hand side.

${(sec x + tan x)}^{2}$ $(secx+tanx)^2$

= ${(\frac{1}{cos x} + \frac{sin x}{cos x})}^{2}$ $(1/cosx+sinx/cosx)^2$

= ${(\frac{1 + sin x}{cos x})}^{2}$ $((1+sinx)/cosx)^2$

= $\frac{{(1 + sin x)}^{2}}{{cos}^{2} x}$ $(1+sinx)^2/cos^2x$

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

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

= $\frac{1 + sin x}{1 - sin x}$ $(1+sinx)/(1-sinx)$

Answer 3 · 2016-10-13T10:31:22

$L H S = \frac{1 + sin x}{1 - sin x}$ $LHS=(1+sinx)/(1-sinx)$

$= \frac{(1 + sin x) (1 + sin x)}{(1 - sin x) (1 + sin x)}$ $=((1+sinx)(1+sinx))/((1-sinx)(1+sinx))$

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

$= \frac{{(1 + sin x)}^{2}}{{cos}^{2} x}$ $=(1+sinx)^2/cos^2x$

$= {(\frac{1 + sin x}{cos x})}^{2}$ $=((1+sinx)/cosx)^2$

$= {(\frac{1}{cos x} + \frac{sin x}{cos x})}^{2}$ $=(1/cosx+sinx/cosx)^2$

$= {(sec x + tan x)}^{2}$ $=(secx+tanx)^2$

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Prove that $\frac{1 + sin x}{1 - sin x} = {(sec x + tan x)}^{2}$ $(1+sinx)/(1-sinx)=(secx+tanx)^2$ ?

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