Math, asked by rajlanjhi6511, 1 year ago

prove that, tan(60+A) tan(60-A) = (2cos2A+1)/(2cos2A -1)

Answers

Answered by ravisanplapcjwiv
67
I hope you will be able to understand
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Answered by aquialaska
28

Answer:

To prove: tan\,(60+A)\:tan\,(60-A)=\frac{2cos\,2A+1}{2cos\,2A-1}

Consider,

LHS

=tan\,(60+A)\:tan\,(60-A)

=(\frac{tan\,60+tan\,A}{1-tan\,60\:tan\,A})(\frac{tan\,60-tan\,A}{1+tan\,60\:tan\,A})

=(\frac{\sqrt{3}+tan\,A}{1-\sqrt{3}\:tan\,A})(\frac{\sqrt{3}-tan\,A}{1+\sqrt{3}\:tan\,A})

=\frac{3-tan^2\,A}{1-3\:tan^2\,A}

RHS

=\frac{2cos\,2A+1}{2cos\,2A-1}

=\frac{2\frac{1-tan^2\,A}{1+tan^2\,A}+1}{2\frac{1-tan^2\,A}{1+tan^2\,A}-1}

=\frac{2-2tan^2\,A+1+tan^1\,A}{1+tan^2\,A}\div\frac{2-2tan^2\,A-1-tan^2\,A}{1+tan^2\,A}

=\frac{3-tan^1\,A}{1+tan^2\,A}\times\frac{1+tan^2\,A}{3-tan^2\,A}

=\frac{3-tan^1\,A}{3-tan^2\,A}

LHS = RHS

Hence Proved.

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