证明以下恒等式:(1+cot A - cosec A)(1+tan A + sec A) = 2
待办事项
我们需要证明 (1+cot A - cosec A)(1+tan A + sec A) = 2。
解答
我们知道:
sin²A + cos²A = 1
cosec²A - cot²A = 1
sec²A - tan²A = 1
cot A = cos A / sin A
tan A = sin A / cos A
cosec A = 1 / sin A
sec A = 1 / cos A
因此:
(1+cot A - cosec A)(1+tan A + sec A) = (1 + cos A / sin A - 1 / sin A)(1 + sin A / cos A + 1 / cos A)
= (sin A + cos A - 1) / sin A) * ((cos A + sin A + 1) / cos A)
= [(sin A + cos A) - 1][(sin A + cos A) + 1] / (sin A cos A)
= [(sin A + cos A)² - 1] / (sin A cos A)
= (sin²A + cos²A + 2sin A cos A - 1) / (sin A cos A)
= (1 + 2sin A cos A - 1) / (sin A cos A)
= 2sin A cos A / (sin A cos A)
$=2$
证毕。广告