Question

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

• A. equal
• B. real
• C. imaginary
• D. greater than $1$

Answer 1

Answer: (B)

real

$\sin a,\sin b,\cos a$ are in G.P.

$\therefore$ ${\sin }^{2}b=\sin a\cos a$ -------- ( 1 )

$\Rightarrow$ $x^2+2x\cot b+1=0$

$\Rightarrow$ $D=(2\cot b{)}^2-4(1\left)\right(1)$

$=4{\cot }^{2}b-1$

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

$=4\left(\frac{{\cos }^{2}b-{\sin }^{2}b}{{\sin }^{2}b}\right)$

$=4\left(\frac{1-{\sin }^{2}b-{\sin }^{2}b}{{\sin }^{2}b}\right)$

$=4\left(\frac{1-2{\sin }^{2}b}{{\sin }^{2}b}\right)$

$=4\left(\frac{1-2\sin a\cos a}{{\sin }^{2}b}\right)$ [ From equation ( 1) ]

$=4\left(\frac{{\sin }^{2}a+{\cos }^{2}a-2\sin a\cos a}{{\sin }^{2}b}\right)$

$=4\left(\frac{(\sin a-\cos b{)}^2}{{\sin }^{2}b}\right)$

$\Rightarrow$ $D={\left(\frac{2(\sin a-\cos a)}{\sin b}\right)}^2\ge 0$

$\therefore$ Roots of given equation are real.