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## QuestionMathsClass 10

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) $$cos\frac{A}{2}$$

(B)
4.6
4.6

## Solution

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