in any triangle ABC show that area of triangle ABC = s(s-a) tanA/2 =s(s-b) tanB/2 =s(s-c) tan C/2 

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Please find below the solution to the asked query:

If a,b,c are sides and A,B,C represents angle corresponding to sides a,b,c, then in ABC, we have:tanA2=s-bs-css-atanB2=s-as-css-btanC2=s-as-bss-cBy Heron's formulaArea=ss-as-bs-c=ss-as-bs-c=ss-as-bs-c=s2s-a2s-bs-css-a=s2s-as-b2s-css-b=s2s-as-bs-c2ss-c=ss-as-bs-css-a=ss-bs-as-css-b=ss-cs-as-bss-c=ss-a.tanA2=ss-b.tanB2=ss-c.tanC2HenceArea of triangle ABC=ss-a.tanA2=ss-b.tanB2=ss-c.tanC2
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