Question

If $\sin \alpha ,\sin \beta$ and $\cos \alpha$ are in $GP$, then roots of $x^2+2x\cot \beta +1=0$ are always.

• A. Real
• B. Real and negative
• C. Greater than one
• D. Non-real

Answer 1

The correct option is A Real

$\sin \alpha ,\sin \beta ,\cos \alpha$ are in G.P.

$\Rightarrow {\sin }^2\beta =\left(\sin \alpha \right)\left(\cos \alpha \right)$

Now, the qudratic is: $x^2+2\cos \beta x+1=0$

$D=b^2-4ac$

$=4{\cot }^2\beta -4\left(1\right)\left(1\right)$

$=4({\cot }^2\beta -1)$

$=4(cose{c}^2\beta -1-1)$

$=4(cose{c}^2\beta -2)$

$=4(\frac{1}{{\sin }^2\beta }-2)$

$=4(\frac{1\times 2}{2\sin \alpha \cos \alpha }-2)$

$=4(\frac{2}{\sin 2\alpha }-2)$

$=4(2cosec2\alpha -2)$

$=8(cosec2\alpha -1)$

At $cosec2\alpha \ge 1$ if $cosec2\alpha$ is positive

$\Rightarrow D\ge 0$

$\Rightarrow$ Roots are real.