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Mathematics

For △ ABC, prove that :

sec A+C2=cosec B2\text{sec }\dfrac{A+C}{2}=\text{cosec }\dfrac{B}{2}

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Answer

In any triangle △ ABC, the sum of all the angles is 180°.

⇒ A + B + C = 180°

⇒ A + C = 180° - B

Dividing both sides by 2:

⇒ A+C2=180°−B2\dfrac{A + C}{2} = \dfrac{180° - B}{2}

⇒ A+C2=180°2−B2\dfrac{A + C}{2} = \dfrac{180°}{2} - \dfrac{B}{2}

⇒ A+C2=90°−B2\dfrac{A + C}{2} = 90° - \dfrac{B}{2}

sec ∠A+∠C2\dfrac{∠A + ∠C}{2} = sec (90°−B2)\Big(90° - \dfrac{B}{2}\Big)

sec ∠A+∠C2\dfrac{∠A + ∠C}{2} = cosec B2\dfrac{B}{2}

Hence, proved sec A+C2=cosec B2\text{sec }\dfrac{A+C}{2} = \text{cosec }\dfrac{B}{2}.

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