In-ABC-holds-2-a-cos-B-2-cos-C-2-s-sec-2B-tan-2B-c-b-c-b-

Question Number 189201 by Shrinava last updated on 13/Mar/23

In △ ABC holds :

\sqrt{2} a \cos \frac{B}{2} \cos \frac{C}{2} = s

\Rightarrow \sec (2 B) + \tan (2 B) = \frac{c + b}{c - b}

Answered by som(math1967) last updated on 13/Mar/23

\sqrt{2} a c o s \frac{B}{2} c o s \frac{C}{2} = s

\sqrt{2} a \sqrt{\frac{s (s - b)}{c a}} \times \sqrt{\frac{s (s - c)}{a b}} = s

\sqrt{2} a \frac{s}{a} \times \sqrt{\frac{(s - b) (s - c)}{b c}} = s

\sqrt{2} s i n \frac{A}{2} = 1

∴ s i n \frac{A}{2} = \frac{1}{\sqrt{2}} \Rightarrow \frac{A}{2} = \frac{π}{4}

∴ A = \frac{π}{2} ∴ a^{2} = b^{2} + c^{2}

n o w s e c 2 B + t a n 2 B

= \frac{1}{c o s 2 B} + \frac{s i n 2 B}{c o s 2 B}

= \frac{1 + s i n 2 B}{c o s 2 B}

= \frac{{(c o s B + s i n B)}^{2}}{{c o s}^{2} B - {s i n}^{2} B}

= \frac{c o s B + s i n B}{c o s B - s i n B}

= \frac{c o s B + s i n (\frac{π}{2} - C)}{c o s B - s i n (\frac{π}{2} - C)} [A = \frac{π}{2} ∴ B + C = \frac{π}{2}]

= \frac{c o s B + c o s C}{c o s B - c o s C}

= \frac{\frac{c^{2} + a^{2} - b^{2}}{2 c a} + \frac{a^{2} + b^{2} - c^{2}}{2 a b}}{\frac{c^{2} + a^{2} - b^{2}}{2 c a} - \frac{a^{2} + b^{2} - c^{2}}{2 a b}}

= \frac{\frac{2 c^{2}}{2 c a} + \frac{2 b^{2}}{2 a b}}{\frac{2 c^{2}}{2 c a} - \frac{2 b^{2}}{2 a b}} [a^{2} = b^{2} + c^{2}]

= \frac{c + b}{c - b}

Commented by Shrinava last updated on 13/Mar/23

thankyou dearSer cool