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avatar+895 

Prove this identity.

\(\frac{1+sin\theta}{1-sin\theta}=(sec\theta+tan\theta)^2\)

From everything I've tried, I don't think this is an identity but maybe I'm wrong.

 May 2, 2019
 #1
avatar+6046 
+2

\(\dfrac{1+\sin(\theta)}{1-\sin(\theta)} \cdot \dfrac{1+\sin(\theta)}{1+\sin(\theta)} = \\ \dfrac{(1+\sin(\theta))^2}{1-\sin^2(\theta)} = \\ \left(\dfrac{1+\sin(\theta)}{\cos(\theta)}\right)^2 = \\ (\sec(\theta)+\tan(\theta))^2\)

.
 May 2, 2019
 #2
avatar+106539 
+2

(sec x + tanx ) ^2  =

 

(1/ cosx   + sinx / cosx)^2  =

 

(1 + sin x)^2          (1+ sin x) (1 + sin x)            (1 + sin x) (1 + sinx)             1 + sin x

_________      =   _________________  =  ___________________ =   ___________

    cos^2x                1 -  sin^2x                         (1 + sin x)(1 - sin x)              1 -   sin x

 

 

 

cool cool cool

 May 3, 2019

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