# If cosec

Question:

If $\operatorname{cosec} x-\sin x=a^{3}$, sec $x-\cos x=b^{3}$, then prove that $a^{2} b^{2}\left(a^{2}+b^{2}\right)=1$

Solution:

$\operatorname{cosec} x-\sin x=a^{3}$

$\therefore \frac{1}{\sin x}-\sin =a^{3}$

$\Rightarrow \frac{1-\sin ^{2} x}{\sin x}=a^{3}$

$\Rightarrow \frac{\cos ^{2} x}{\sin x}=a^{3}$

$\Rightarrow a=\left(\frac{\cos ^{2} x}{\sin x}\right)^{\frac{1}{3}} \quad \ldots(\mathrm{i})$

Also, sec $x-\cos x=b^{3}$

$\Rightarrow \frac{1}{\cos x}-\cos =b^{3}$

$\Rightarrow \frac{1-\cos ^{2} x}{\cos x}=b^{3}$

$\Rightarrow \frac{\sin ^{2} x}{\cos x}=b^{3}$

$\Rightarrow b=\left(\frac{\sin ^{2} x}{\cos x}\right)^{\frac{1}{3}} \quad \ldots . .(\mathrm{ii})$

Now, LHS $=a^{2} b^{2}\left(a^{2}+b^{2}\right)=(a b)^{2}\left(a^{2}+b^{2}\right)$

$=\left[\left(\frac{\cos ^{2} x}{\sin x}\right)^{\frac{1}{3}}\left(\frac{\sin ^{2} x}{\cos x}\right)^{\frac{1}{3}}\right]^{2}\left[\left(\left(\frac{\cos ^{2} x}{\sin x}\right)^{\frac{1}{3}}\right)^{2}+\left(\left(\frac{\sin ^{2} x}{\cos x}\right)^{\frac{1}{3}}\right)^{2}\right]$

$=(\sin x \cos x)^{\frac{2}{3}}\left[\frac{\left(\cos ^{2} x\right)^{\frac{2}{3}}}{(\sin x)^{\frac{2}{3}}}+\frac{\left(\sin ^{2} x\right)^{\frac{2}{3}}}{(\cos x)^{\frac{2}{3}}}\right]$

$=(\sin x \cos x)^{\frac{2}{3}}\left[\frac{\left(\cos ^{3} x\right)^{\frac{2}{3}}+\left(\sin ^{3} x\right)^{\frac{2}{3}}}{(\sin x)^{\frac{2}{3}}(\cos x)^{\frac{2}{3}}}\right]$

$=(\sin x \cos x)^{\frac{2}{3}}\left[\frac{\cos ^{2} x+\sin ^{2} x}{(\sin x \cos x)^{\frac{2}{3}}}\right]$

= 1= RHS