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

Posted by Chirag Bhardwaj 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

We know that, A + B + C = 180^o
{tex}\Rightarrow{/tex} A + B = 180^o - C
Dividing both sides with 2, we get
{tex}\Rightarrow{/tex} {tex}\frac{A + B}{2}{/tex} = 90^o - {tex}\frac{C}{2}{/tex}
Applying cosec on both sides, we get
{tex}\Rightarrow{/tex} cosec({tex}\frac{A + B}{2}{/tex}) = cosec(90^o - {tex}\frac{C}{2}{/tex})
{tex}\Rightarrow{/tex} cosec({tex}\frac{A + B}{2}{/tex}) = sec {tex}\frac{C}{2}{/tex}

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CBSE > Class 10 > Mathematics