Question

if a b c are the angles of a triangle such that sec(a-b), seca , sec(a+b) are in AP then show that 2Sec^2a=sec^2b/2​

Answer 1

2sec a=sec (a+b)+sec(a-b) {as in ap}

$2seca = sec(a + b) + sec(a - b) \\2sec a= \frac{ \cos(a - b) + \cos(a + b) }{ \cos(a). \cos(b) - \sin(a). \sin(b) - \cos(a). \cos(b) + \sin(a). \sin(b) } \\ \\ 2seca = \frac{2 \cos(a) . \cos(b) }{ { cos}^{2}a. {cos}^{2} b - {sin}^{2} a. {sin}^{2} b} \\ \frac{2}{cosa} = \frac{2 \cos(a) . \cos(b)}{ {cos}^{2} b + {cos}^{2} a - 1} \\ {{cos}^{2} b + {cos}^{2} a - 1} = {cos}^{2} a.cos(b)$

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Answer 2

Step-by-step explanation:

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