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Question and Answer

If A, B, C are the interior angles of a triangles, then \( tan\left(\frac{B+C}{2}\right)=\)
(A) \( tan\frac{A}{2}\)
(B) \( cot\frac{A}{2}\)
(C) \( sin\frac{A}{2}\)
(D) \( cos\frac{A}{2}\)

Answer

(B)
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Solution

\( A+B+C=180°\)
\( B+C=180°A\)
\( \frac{B+C}{2}=\frac{180°-A}{2}\)
\( \frac{B+C}{2}=90°-\frac{A}{2}\)
\( tan\left(\frac{B+C}{2}\right)=tan\left(90°-\frac{A}{2}\right)\)
\( cot\frac{A}{2}\)
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