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CGPET 2015 Solved Paper

Section: Mathematics

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Question : 141 of 150

Marks: +1, -0

In

\triangle ABC

\cot A, \cot B

\cot C

are in AP, then

a^2, b^2

c^2

HP
AP
GP
None of the above

Solution:

It is given, cot A, cot B and cot C are in AP.

⇒

2\cot B = \cot A + \cot C

⇒

2\frac{\cos B}{\sin B} = \frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}

∵

\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k

⇒

\frac{2\cos B}{kb} = \frac{\cos A}{ak} + \frac{\cos C}{ck}

⇒

\frac{2\cos B}{b} = \frac{\cos A}{a} + \frac{\cos C}{c}

⇒

\frac{2}{b} \left[ \frac{a^2+c^2-b^2}{2ac} \right] = \frac{1}{a} \left[ \frac{b^2+c^2-a^2}{2bc} + \frac{1}{c} \left[ \frac{a^2+b^2-c^2}{2ab} \right] \right]

⇒

\frac{2(a^2+c^2-b^2)}{2abc} = \frac{1}{2abc} \left[ (b^2+c^2-a^2)+(a^2+b^2-c^2) \right]

⇒

2(a^2+c^2-b^2)=2b^2

⇒

a^2+c^2=2b^2

∴

a^2, b^2 \text{ and } c^2 \text{ are in AP.}

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