Question

If $\alpha ,\beta ,\gamma$ be the three angles given by $\cos \alpha =\frac{a}{b+c},\cos \beta =\frac{b}{c+a}$ and $\cos \gamma =\frac{c}{a+b}$ where $a,b,c$ are the sides of a $△ABC$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$ is equal to

• A. $\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
• B. $\csc \frac{A}{2}\csc \frac{B}{2}\csc \frac{C}{2}$
• C. $\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$
• D. $1$

Answer 1

Answer: (C)

$\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$

Given

$\cos \alpha =\frac{a}{b+c}$

$⟹\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}=\frac{a}{b+c}$.......................... $(\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta })$

$⟹(b+c)-(b+c){\tan }^2\frac{\alpha }{2}=a+a{\tan }^2\frac{\alpha }{2}$

$⟹(b+c)-a=a{\tan }^2\frac{\alpha }{2}+(b+c){\tan }^2\frac{\alpha }{2}$

$⟹(b+c-a)=(a+b+c){\tan }^2\frac{\alpha }{2}$

$⟹{\tan }^2\frac{\alpha }{2}=\frac{b+c-a}{a+b+c}$

$⟹{\tan }^2\frac{\alpha }{2}=\frac{2s-2a}{2s}$.................. $(s=\frac{a+b+c}{2})$

$⟹{\tan }^2\frac{\alpha }{2}=\frac{s-a}{s}\cdots \left(1\right)$

Similarly, ${\tan }^2\frac{\beta }{2}=\frac{s-b}{s}\cdots \left(2\right)$

${\tan }^2\frac{\gamma }{2}=\frac{s-c}{s}\cdots \left(3\right)$

From $\left(1\right),\left(2\right),\left(3\right)$

$⟹{\tan }^2\frac{\alpha }{2}{\tan }^2\frac{\beta }{2}{\tan }^2\frac{\gamma }{2}=\left(\frac{s-a}{s}\right)\left(\frac{s-b}{s}\right)\left(\frac{s-c}{s}\right)$

Multiplying and Diving by $s$

$⟹\frac{s(s-a)(s-b)(s-c)}{s^4}$

$⟹\frac{{\Delta }^2}{s^4}\cdots \left(4\right)$

$(\Delta =\sqrt{s(s-a)(s-b)(s-c)})$

Considering $\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$

$⟹\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}=\left(\frac{(s-b)(s-c)}{s(s-a)}\right)\left(\frac{(s-a)(s-c)}{s(s-b)}\right)\left(\frac{(s-a)(s-b)}{s(s-c)}\right)$

$⟹\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}=\left(\frac{(s-a{)}^2(s-b{)}^2(s-c{)}^2}{s^3(s-a)(s-b)(s-c)}\right)$

$⟹\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}=\left(\frac{(s-a)(s-b)(s-c)}{s^3}\right)$

Multiplying and Diving by $s$

$⟹\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}=\left(\frac{s(s-a)(s-b)(s-c)}{s^4}\right)$

$⟹\frac{{\Delta }^2}{s^4}\cdots \left(5\right)$

$(\Delta =\sqrt{s(s-a)(s-b)(s-c)})$

From $\left(5\right),\left(6\right)$

$⟹{\tan }^2\frac{\alpha }{2}{\tan }^2\frac{\beta }{2}{\tan }^2\frac{\gamma }{2}=\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$