Respuesta :

Hello,

We have the equation:

( Cosx + Cosx.tg^2x).Secx

We know too:

Tgx = Senx/Cosx

Then,

Cosx.Tg^2x = Cosx . Sen^2x / cos^2x

= Sen^2x/cosx

Then, we stay with:

(Cosx + Sen^2x/Cosx) . Secx

There is a trigonometric property that says:

Sec x = 1 / Cosx

Then, this mean:

Secx = 1 / cos x

Then,

( Cosx + Sen^2/Cosx). 1/Cosx

Applying the distributive:

Cosx . 1/Cosx + Sen^2x / Cos^2x

Simplifying Cos x/ cosx

1 + Sen^2x / Cos^2x

Rewrinting Sen^2x/Cos^2x = Tg^2x

Then us stay:

= 1 + Tg^2x = Sec^2x <=> I hope this has helped
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