How do you show that $cosx/(1 - sinx) = secx+ tanx$ ?

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Answer 1 · 2016-12-04T19:11:47

We know that $sectheta = 1/costheta$ and $tantheta= sintheta/costheta$ .

$cosx/(1 - sinx) = 1/cosx + sinx/cosx$

$cosx/(1- sinx) = (1 + sinx)/cosx$

Multiply the left side by the conjugate of the denominator. The conjugate of $a + b$ is $a- b$ , for example.

$cosx/(1 - sinx) xx (1 + sinx)/(1 + sinx) = (1 + sinx)/cosx$

$(cosx + cosxsinx)/(1 - sin^2x) = (1 + sinx)/cosx$

Use the identity $sin^2theta + cos^2theta= 1-> cos^2theta = 1- sin^2theta$ .

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

$(1 + sinx)/cosx = (1 + sinx)/cosx$

$LHS = RHS$

Identity Proved!

Hopefully this helps!

How do you show that cosx/(1 - sinx) = secx+ tanxcosx/(1 - sinx) = secx+ tanx?