IIT JEE , AIEEE Forum - Mathematics , Algebra

waterdemon

Here is bit different method.

We know that Sin , Sin , Cos are in GP

so we can say that Sin² = Sin *Cos

Now from the quadratic equation x² + 2xCot + 1

D = 4Cot² - 4 = 4 ( Cot² - 1 )

Now we can write Sin² = Sin2/2

Therefore Sin =

Sin2/2

Now take a triangle and an angle .

From there we will get Cot = 2-Sin2/Sin2

Therefore : Cot² = 2-Sin2/Sin2

Now putting the value above in the discriminant.

D = 4 (( 2-Sin2/Sin2 ) - 1)

D = 4 ((2/Sin2- 1) + 1)

D = 4 (2/Sin2)

D = 8/Sin2

We know that Sin2 will always be positive . and 8 is positive.

So we will get the whole term as that is

D > 0

Hence the roots of the quadratic eqaution are "Real".

Hence option (b) is right.

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