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Please help! 

 

\(tan(x+y)tan(x-y)=\frac{sin^2x-sin^2y}{cos^2x-sin^2y}\)

 

and this one: 

\(\frac{1+cosx}{sinx}=cot\left(\frac{x}{2}\right)\)

 

 

Thank you so much!! 

 Apr 12, 2020
 #1
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for the second 1 multiply top and bottom by 1-cos x

 Apr 12, 2020
 #2
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For:  tan(x + y)tan(x - y)  =  ( sin2x - sin2y ) / ( cos2x - sin2y )

                                       =  [ (sinx + siny)(sinx - siny) ] / [ (cosx + cosy)(cosx - cosy) ]

                                       =  [ (sinx + siny) / (cosx + cosy) ] · [ (sinx - siny) / (cosx - cosy) ]

                                       =  tan(x + y) · tan(x - y)

 Apr 13, 2020

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