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If A + B + C = 180°. Prove that cot B. cot C + cot C · cot A + cot A · cot B = 1 |
Answer» Given A + B + C = 180°; A + B = 180° – C; cot(A + B) = cot (180° – C) \(\frac{cot(A).cot(B) - 1}{cotA + cotB} = - cotC\) cotA . cotB – 1 = -cotA · cotC – cot B . cotC. ∴ cotA . cotB + cotB cotC + cotC . cotA = 1. |
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