Sec thita -1÷ sec thita +1 =(sin thita ÷ 1+ costhita)^2 ?

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Sec thita -1÷ sec thita +1 =(sin thita ÷ 1+ costhita)^2 ?

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Answer 1 · 2018-05-22T09:48:26

Please see the proof below

Explanation:

We need

$sec θ = \frac{1}{cos θ}$ $sectheta=1/costheta$

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

Therefore the

$L H S = \frac{sec θ - 1}{sec θ + 1}$ $LHS=(sectheta-1)/(sectheta+1)$

$= \frac{\frac{1}{cos θ} - 1}{\frac{1}{cos θ} + 1}$ $=(1/costheta-1)/(1/costheta+1)$

$= \frac{1 - cos θ}{1 + cos θ}$ $=(1-costheta)/(1+costheta)$

$= \frac{(1 - cos θ) (1 + cos θ)}{(1 + cos θ) (1 + cos θ)}$ $=((1-costheta)(1+costheta))/((1+costheta)(1+costheta))$

$= \frac{1 - {cos}^{2} θ}{{(1 + cos θ)}^{2}}$ $=(1-cos^2theta)/(1+costheta)^2$

$\frac{{sin}^{2} θ}{{(1 + cos θ)}^{2}}$ $sin^2theta/(1+costheta)^2$

$= {(\frac{sin θ}{1 + cos θ})}^{2}$ $=(sintheta/(1+costheta))^2$

$= R H S$ $=RHS$

$Q E D$ $QED$

Answer 2 · 2018-05-22T10:41:05

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

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

$= \frac{1 - cos x}{1 + cos x} \cdot [\frac{1 + cos x}{1 + cos x}]$ $=(1-cosx)/(1+cosx)*[(1+cosx)/(1+cosx)]$

$= \frac{1 - {cos}^{2} x}{{(1 + cos x)}^{2}} = \frac{{sin}^{2} x}{{(1 + cos x)}^{2}} = {(\frac{sin x}{1 + cos x})}^{2} = R H S$ $=(1-cos^2x)/(1+cosx)^2=sin^2x/(1+cosx)^2=(sinx/(1+cosx))^2=RHS$

Answer 3 · 2018-05-22T10:44:59

Explanation in below

Explanation:

$\frac{sec x - 1}{sec x + 1}$ $(secx-1)/(secx+1)$

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

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

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

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

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

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

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

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Sec thita -1÷ sec thita +1 =(sin thita ÷ 1+ costhita)^2 ?

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