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A,B,C are interior angles of ∆ABC …

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A,B,C are interior angles of ∆ABC . Prove that cosec(A+B/2) = sec C/2.

Posted by Dipali Shrivastava 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Rakesh Kumar Tiwari 6 years, 10 months ago

cosec(A+B)/2 = cosec(180-C)/2. [since A+B+C=180] = cosec (180/2 - C/2) = cosec (90 - C/2) = sec C/2

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