Question

In a $\Delta ABC,A>C,$ $A-C={90}^0,$ $a+c=2b$ and if $p=\sin \frac{B}{2}$ , $q=\cos \frac{B}{2}$ , $r=\tan \frac{B}{2}$ then the ascending order of $p,q,r$ is?

• A. $p<q<r$
• B. $p<r<q$
• C. $r<p<q$
• D. $q<p<r$

Answer 1

The correct option is B $p<r<q$

$A-C={90}^o,$ $a+c=2b$

$p=\sin \frac{B}{2},q=\cos \frac{B}{2},r=\tan \frac{B}{2}$

$\frac{\sin \left(\frac{A-C}{2}\right)}{\sin \left(\frac{A+C}{2}\right)}=\frac{\tan \left(\frac{a-c}{2}\right)}{\tan \left(\frac{c}{2}\right)}$

$\frac{\frac{1}{\sqrt{2}}}{\cos \left(\frac{B}{2}\right)}=\frac{\tan \left(\frac{a-c}{2}\right)}{\tan \left(\frac{c}{2}\right)}$

$\frac{\frac{1}{\sqrt{2}}}{q}=\frac{\tan \left(\frac{a-c}{2}\right)}{\tan \left(\frac{c}{2}\right)}$

$q=\frac{\frac{1}{\sqrt{2}}\tan \left(\frac{c}{2}\right)}{\tan \left(\frac{a-c}{2}\right)}$

$\frac{\cos \left(\frac{A-C}{2}\right)}{\cos \left(\frac{A+C}{2}\right)}=\frac{\tan \left(\frac{a+c}{2}\right)}{\tan \left(\frac{c}{2}\right)}$

$\frac{\frac{1}{\sqrt{2}}}{P}=\frac{\tan b}{\tan \left(\frac{c}{2}\right)}$

$P=\frac{\frac{1}{\sqrt{2}}\tan \left(\frac{c}{2}\right)}{\tan b}$

$\frac{\tan \left(\frac{A-C}{2}\right)}{\cot \left(\frac{c}{2}\right)}=\frac{\cos \left(\frac{a-c}{2}\right)}{\cos \left(\frac{a+c}{2}\right)}$

$\frac{1}{\cot \left(\frac{c}{2}\right)}=\frac{\frac{1}{\sqrt{2}}}{P},p=\frac{1}{\sqrt{2}\tan \frac{c}{2}}$

$p=\frac{1}{\sqrt{2}\tan \left(\frac{c}{2}\right)}$

$q=\frac{\frac{1}{\sqrt{2}}\tan \left(\frac{c}{2}\right)}{\tan \left(\frac{a-c}{2}\right)}$

$\tan b=\frac{\frac{1}{\sqrt{2}}\tan \left(\frac{c}{2}\right)}{P}$

$\therefore p<r<q$

Hence, $p<r<q$ is the correct answer.