In ABC- prove that a 2 - b-c 2 -4bcsin 2 A-2 and hence show that a-b-c- where tan- - 2- bc sin A-2 b-c-

R.H.S=( b c ) ^ 2 +4bcsin ^ 2 ( A/2 ) #160; =( b c ) ^ 2 +2bc( 1 cos A #160; ) =b^ 2 +c^ 2 2abcos A =a^ 2 Now squaring the second relation ...

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Principal Solution of Trigonometric Equation

In ABC, pro...

Question

In $△ A B C$ , prove that $a^{2} = {(b - c)}^{2} + 4 b c {sin}^{2} (A / 2)$
and hence show that $a = (b - c) sec ϕ$ where
$tan ϕ = \frac{2 \sqrt{b c} sin (A / 2)}{b - c}$ .

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△ A B C,

2 a c sin (\frac{A - B + C}{2}) = a^{2} + c^{2} - b^{2}

\frac{cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2}}{cot A + cot B + cot C} = \frac{{(a + b + c)}^{2}}{a^{2} + b^{2} + c^{2}}

\frac{b + c}{a} \leq csc \frac{A}{2}

{csc}^{n} \frac{A}{2} {csc}^{n} \frac{B}{2} {csc}^{n} \frac{C}{2} \geq 2^{3 n}

n \geq 1

\frac{{cos}^{2} \frac{B - C}{2}}{(b + c)^{2}} + \frac{{sin}^{2} \frac{B - C}{2}}{(b - c)^{2}} = \frac{1}{a^{2}}

a^{2} + c^{2} = 2 b^{2}

\frac{s i n 3 B}{s i n B} = [\frac{a^{2} - c^{2}}{2 a c}]^{2}

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