With usual notations, in A B C, if 2 a²=b²+c², then 3 A/ A+2=

Correct Answer is : (b2−c22bc)2≤ft({b²-c²}{2 b c})²(2bcb2−c2)2

Correct Answer is : (b2−c22bc)2≤ft({b²-c²}{2 b c})²(2bcb2−c2)2

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MHTCET 13 Apr 2026 Shift 1 Paper

Question : 114 of 150

Marks: +1, -0

\triangle A B C

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

\frac{\cos 3 A}{\cos A}+2=

Solution:

Concept:

Use the cosine rule to find

\cos A

, then apply the triple-angle formula.

Explanation:

Given

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

, so

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

By the cosine rule,

a^{2}=b^{2}+c^{2}-2bc\cos A

Substituting,

\frac{b^{2}+c^{2}}{2}=b^{2}+c^{2}-2bc\cos A

This gives

2bc\cos A=\frac{b^{2}+c^{2}}{2}

Therefore,

\cos A=\frac{b^{2}+c^{2}}{4bc}

Now use

\cos 3A=4\cos^{3}A-3\cos A

Dividing by

\cos A

\frac{\cos 3A}{\cos A}=4\cos^{2}A-3

Hence,

\frac{\cos 3A}{\cos A}+2=4\cos^{2}A-1

Substitute

\cos A=\frac{b^{2}+c^{2}}{4bc}

4\cos^{2}A-1=4\left(\frac{b^{2}+c^{2}}{4bc}\right)^{2}-1

=\frac{(b^{2}+c^{2})^{2}}{4b^{2}c^{2}}-1

=\frac{(b^{2}+c^{2})^{2}-4b^{2}c^{2}}{4b^{2}c^{2}}

=\frac{(b^{2}-c^{2})^{2}}{4b^{2}c^{2}}

=\left(\frac{b^{2}-c^{2}}{2bc}\right)^{2}

Answer:

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

, i.e. Option B.

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