How do you prove $sec x + sin x = tan x sin x$ ?

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Answer 1 · 2016-04-26T10:43:39

This is true only for $x=-\pi/4$ plus multiples of $\pi$ .

Since $\sec(x)=1/\cos(x)$ and $\tan(x)=\sin(x)/\cos(x)$ , first multiply by $\cos(x)$ . Thus

$1+\sin(x)\cos(x)=\sin^2(x)$

Put in #1=\sin^2(x)+\cos^2(x):

$\sin^2(x)+\cos^2(x)+\sin(x)\cos(x)=\sin^2(x)$

$\cos^2(x)+\sin(x)\cos(x)=0$

$\cos(x)(\cos(x)+\sin(x))=0$

Thus the equation is true only if $\cos(x)=0$ or $\cos(x)=-sin(x)$ .

But we cannot accept $\cos(x)=0$ because that leaves $\sec(x)=1/\cos(x)$ undefined. So $\cos(x)=-sin(x)$ , thus $x=-\pi/4$ plus multiples of $\pi$ .

How do you prove sec x + sin x = tan x sin xsec x + sin x = tan x sin x?