Question

In a $\Delta$ ABC if b = 3, c = 5 and cos(B - C) = $\frac{7}{25}$, then find the value of $tan\frac{A}{2}$

• A. 1
• B. $\frac{1}{3}$
• C. $\frac{1}{5}$
• D. $\frac{1}{7}$

Answer 1

The correct option is B $\frac{1}{3}$

Well, here in the problem we can’t directly apply Napier’s analogy as we don’t know what $tan\left(\frac{B-C}{2}\right)$ is. If we know that we can proceed.
Well, if you remember we studied a formula of Tan, Sin and Cos half angle formula. If not, you may need to revise.
$tan\left(\frac{\theta }{2}\right)=\sqrt{\frac{1-cos\theta }{1-cos\theta }}$
So,
$tan\left(\frac{B-C}{2}\right)=\sqrt{\frac{1-cos(B-C)}{1+cos(B-C)}}$
$tan\left(\frac{B-C}{2}\right)=\sqrt{\frac{1-\frac{7}{25}}{1+\frac{7}{25}}}$
$tan\left(\frac{B-C}{2}\right)=\sqrt{\frac{18}{32}}$
$tan\left(\frac{B-C}{2}\right)=\sqrt{\frac{9}{16}}$
$tan\left(\frac{B-C}{2}\right)=\frac{3}{4}$
Now we can apply Napier's Analogy -
$tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}cot\frac{A}{2}$
$\frac{3}{4}=\frac{3-5}{3+5}cot\frac{A}{2}$
$cot\frac{A}{2}=3$
$tan\frac{A}{2}=\frac{1}{3}$