Question #d07bf

1 Answer
Nov 27, 2017

Please see below.

Explanation:

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We want to prove:

#tan(s+t)tan(s-t)=(tan^2s-tan^2t)/(1-tan^2stan^2t)#

From the Sum and Difference formulas we have:

#tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)#

and

#tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta#

Using these identities we get:

#tan(s+t)tan(s-t)=((tans+tant)/(1-tanstant))((tans-tant)/(1+tanstant))=(tan^2s-tan^2t)/(1-tan^2stan^2t)#