A,b,c are interior angle of triangle abc. Prove that cosec(A+B)÷2 =se

1.	A,b,c are interior angle of triangle abc. Prove that cosec(A+B)÷2 =sec c÷2
Answer» We know that, A\xa0+ B + C = 180o\xa0{tex}\\Rightarrow{/tex}\xa0A + B =\xa0180o - CDividing both sides with 2, we get{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{A + B}{2}{/tex}\xa0= 90o -\xa0{tex}\\frac{C}{2}{/tex}Applying cosec on both sides, we get{tex}\\Rightarrow{/tex}\xa0cosec({tex}\\frac{A + B}{2}{/tex}) = cosec(90o -\xa0{tex}\\frac{C}{2}{/tex}){tex}\\Rightarrow{/tex}\xa0cosec({tex}\\frac{A + B}{2}{/tex}) = sec\xa0{tex}\\frac{C}{2}{/tex}

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