With usual notations, in triangle ABC, if 2a^ | vedclass

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(B) Given $2a^2 = b^2 + c^2$. By the Law of Cosines, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting $b^2 + c^2 = 2a^2$, we get $\cos A = \frac{2a

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$(\frac{b^2 - c^2}{2bc})^2$

Vedclass · Answer

(B) Given $2a^2 = b^2 + c^2$. By the Law of Cosines, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting $b^2 + c^2 = 2a^2$, we get $\cos A = \frac{2a^2 - a^2}{2bc} = \frac{a^2}{2bc}$.Using the identity $\cos 3A = 4\cos^3 A - 3\cos A$, we have $\frac{\cos 3A}{\cos A} = 4\cos^2 A - 3$.Thus, $\frac{\cos 3A}{\cos A} + 2 = 4\cos^2 A - 3 + 2 = 4\cos^2 A - 1$.Substitute $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Since $a^2 = \frac{b^2 + c^2}{2}$, $\cos A = \frac{b^2 + c^2 - (b^2 + c^2)/2}{2bc} = \frac{b^2 + c^2}{4bc}$.Then $4\cos^2 A - 1 = 4(\frac{b^2 + c^2}{4bc})^2 - 1 = \frac{(b^2 + c^2)^2}{4b^2c^2} - 1 = \frac{b^4 + c^4 + 2b^2c^2 - 4b^2c^2}{4b^2c^2} = \frac{b^4 + c^4 - 2b^2c^2}{4b^2c^2} = \frac{(b^2 - c^2)^2}{4b^2c^2} = (\frac{b^2 - c^2}{2bc})^2$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

With usual notations, in $\triangle ABC$, if $2a^2 = b^2 + c^2$, then $\frac{\cos 3A}{\cos A} + 2 = $

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