If cos A + cos B + 2cos C = 2,then the sides | vedclass

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(A) Given $\cos A + \cos B + 2\cos C = 2$.Using $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ and $2\cos C = 2(1 - 2\sin^2 \frac{C}{2})

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$A.P.$

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(A) Given $\cos A + \cos B + 2\cos C = 2$.Using $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ and $2\cos C = 2(1 - 2\sin^2 \frac{C}{2}) = 2 - 4\sin^2 \frac{C}{2}$.Substituting these into the equation: $2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} + 2 - 4\sin^2 \frac{C}{2} = 2$.Since $\frac{A+B}{2} = 90^\circ - \frac{C}{2}$,we have $\cos \frac{A+B}{2} = \sin \frac{C}{2}$.So,$2 \sin \frac{C}{2} \cos \frac{A-B}{2} - 4\sin^2 \frac{C}{2} = 0$.$2 \sin \frac{C}{2} (\cos \frac{A-B}{2} - 2\sin \frac{C}{2}) = 0$.Since $\sin \frac{C}{2} 
eq 0$,we have $\cos \frac{A-B}{2} = 2\sin \frac{C}{2} = 2\cos \frac{A+B}{2}$.Using the product-to-sum formula,this simplifies to $\sin A + \sin B = 2\sin C$.By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a + b = 2c$.Thus,the sides $a, c, b$ are in $A.P.$

If $\cos A + \cos B + 2\cos C = 2$,then the sides of the $\Delta ABC$ are in

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