Question

If $p=\sqrt{\frac{1-\sin x}{1+\sin x}},q=\frac{1-\sin x}{\cos x},r=\frac{\cos x}{1+\sin x}$, which is the correct statement?

• A. $p=q\ne r$
• B. $q=r\ne p$
• C. $r=p\ne q$
• D. $p=q=r$

Answer 1

The correct option is D $p=q=r$

$p=\sqrt{\frac{1-\sin x}{1+\sin x}}$
$p=\sqrt{\frac{1-2\sin \frac{x}{2}\cos \frac{x}{2}}{1+2\sin \frac{x}{2}\cos \frac{x}{2}}}[\because \sin x=2\sin \frac{x}{2}\cos \frac{x}{2}]$
$p=\sqrt{\frac{{\sin }^{2}x/2+{\cos }^{2}x/2-2\sin x/2\cos x/2}{{\sin }^{2}x/2+{\cos }^{2}x/2+2\sin x/2\cos x/2}}[\because {\sin }^{2}x/2+{\cos }^{2}x/2=1]$
$p=\sqrt{\frac{{(\cos x/2-\sin x/2)}^2}{{(\sin x/2+\cos x/2)}^2}}$
$p=\frac{\cos x/2-\sin x/2}{\sin x/2+\cos x/2}⟶1$
$q=\frac{1-\sin x}{\cos x}=\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}{{\cos }^{2}x/2-{\sin }^{2}x/2}$
$=\frac{{[\cos x/2-\sin x/2]}^2}{[\cos x/2-\sin x/2][\cos x/2+\sin x/2]}$
$q=\frac{\cos x/2-\sin x/2}{\cos x/2+\sin x/2}⟶2$
$r=\frac{\cos x}{1+\sin x}=\frac{{\cos }^{2}x/2-{\sin }^{2}x/2}{{\cos }^{2}x/2+{\sin }^{2}x/2+2\sin x/2\cos x/2}$
$=\frac{[\cos x/2-\sin x/2][\cos x/2+\sin x/2]}{{[\sin x/2+\cos x/2]}^2}$
$r=\frac{\cos x/2-\sin x/2}{\cos x/2+\sin x/2}⟶2$
From $1,2$ & $3\Rightarrow p=q=r$