Question

If sin^3x + cosec^3x = 110, then the value of sin^2x + cosec^2x is equal to

Answer 1

Let $\sin x=k$. Then $\csc x=\dfrac {1}{k}.$ So we have the equation,

$k^3+\dfrac {1}{k^3}=110\quad\longrightarrow\quad (1)$

But we know that,

$a^3+b^3=(a+b)(a^2-ab+b^2)$

Then,

$\left (k+\dfrac {1}{k}\right)\left (k^2+\dfrac {1}{k^2}-1\right)=110\\\\\\\left (k+\dfrac {1}{k}\right)\left (k^2+\dfrac {1}{k^2}+2-3\right)=110\\\\\\\left (k+\dfrac {1}{k}\right)\left (\left (k+\dfrac {1}{k}\right)^2-3\right)=110$

Let $k+\dfrac {1}{k}=p.$ Then,

$p(p^2-3)=110\\\\p^3-3p-110=0\\\\p^3-5p^2+5p^2-25p+22p-110=0\\\\p^2(p-5)+5p(p-5)+22(p-5)=0\\\\(p-5)(p^2+5p+22)=0$

Since the roots of $p^2+5p+22=0$ are not real,

$p=5\\\\\\k+\dfrac {1}{k}=5\\\\\\\left (k+\dfrac {1}{k}\right)^2=5^2\\\\\\k^2+\dfrac {1}{k^2}+2=25\\\\\\k^2+\dfrac {1}{k^2}=23$

That is,

$\large\boxed {\sin^2x+\csc^2x=\mathbf {23}}$

#answerwithquality

#BAL