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Mathematics 17 Online
OpenStudy (anonymous):

How do you simplify (Cosx)/(Secx)+(Tanx) ?

OpenStudy (anonymous):

is the question like this? \[\frac{ \cos x }{\sec x+\tan x }\]

OpenStudy (anonymous):

Yes Exactly.

OpenStudy (anonymous):

\[\frac{ \cos x }{\sec x+\tan x }\times \frac{ \sec x-\tan x }{\sec x-\tan x }=\frac{\cos x \left( \sec x-\tan x \right) }{ \sec ^2 x -\tan ^2x }\] \[\left[ \sec ^2x-\tan ^2x=1 \right]\] simplify it.

OpenStudy (unklerhaukus):

Alternately: \[\frac{ \cos x }{\sec x+\tan x }= \frac{ \cos x }{\dfrac1{\cos x}+\dfrac{\sin x}{\cos x} }\\\qquad\qquad\qquad=\frac{\cos^2x}{1+\sin x}\] then use \(\sin^2 x+ \cos^2 x =1\\ \qquad\qquad\cos^2 x=1-\sin^2 x\)

OpenStudy (unklerhaukus):

and \(1-z^2=(1-z)(1+z)\)

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