Math, asked by leoHarshit6174, 1 year ago

s(s-a)tanA/2=s(s-b)tanB/2=s(s-c)tanC/2

Answers

Answered by jitumahi435
1

To prove that:

s(s-a)\tan \dfrac{A}{2} =s(s-b)\tan \dfrac{B}{2}=s(s-c)\tan \dfrac{C}{2}.

L.H.S. = s(s-a)\tan \dfrac{A}{2}

Using the identity:

\tan \dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)} }

= s(s-a) \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)} }

=\sqrt{s(s-a)(s-b)(s-c)

= Δ

M.H.S. = s(s-b)\tan \dfrac{B}{2}

Using the identity:

\tan \dfrac{B}{2}=\sqrt{\dfrac{(s-c)(s-a)}{s(s-b)} }

= s(s-b) \sqrt{\dfrac{(s-c)(s-a)}{s(s-b)} }

=\sqrt{s(s-a)(s-b)(s-c)

= Δ

R.H.S. = s(s-c)\tan \dfrac{C}{2}

Using the identity:

\tan \dfrac{C}{2}=\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)} }

= s(s-c) \sqrt{\dfrac{(s-a)(s-b)}{s(s-c)} }

=\sqrt{s(s-a)(s-b)(s-c)

= Δ

∴ L.H.S. = M.H.S. = M.H.S. = Δ

Thus, s(s-a)\tan \dfrac{A}{2} =s(s-b)\tan \dfrac{B}{2}=s(s-c)\tan \dfrac{C}{2}, proved.

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