How do you prove $2tan(x)sec(x) = (cos^2x+sin^2x)/(1-sin(x)) - (1/(1+sin(x)))$ ?

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Answer 1 · 2018-03-06T04:39:34

Please refer to a Proof in the Explanation.

Since, $cos^2x+sin^2x=1$ , we have,

$(cos^2x+sin^2x)/(1-sinx)-1/(1+sinx)$ ,

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

$={1(1+sinx)-1(1-sinx)}/{(1-sinx)(1+sinx)}$ ,

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

$=(2sinx)/cos^2x$ ,

$=2*sinx/cosx*1/cosx$ ,

$=2tanxsecx,$ as desired!

How do you prove 2tan(x)sec(x) = (cos^2x+sin^2x)/(1-sin(x)) - (1/(1+sin(x))) 2tan(x)sec(x) = (cos^2x+sin^2x)/(1-sin(x)) - (1/(1+sin(x))) ?