1
$\begingroup$

If in a triangle ABC, $b+c=3a$, then $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ is equal to ?

My reference gives the solution $2$, but I have no clue of where to start ?

My Attempt $$ \cot\dfrac{B}{2}.\cot\dfrac{C}{2}=\frac{\cos\frac{B}{2}.\cos\frac{C}{2}}{\sin\frac{B}{2}.\sin\frac{C}{2}}=\frac{\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})}{\cos(\frac{A-B}{2})-\cos(\frac{A+B}{2})} $$

$\endgroup$
2
$\begingroup$

Hint:

$$b+c=3a\implies\sin B+\sin C=3\sin A$$

$$\iff2\sin\dfrac{B+C}2\cos\dfrac{B-C}2=6\sin\dfrac A2\cos\dfrac A2$$

Now use $\dfrac{B+C}2=\dfrac\pi2-\dfrac A2,\cos\dfrac{B+C}2=?$

As $0<A<\pi,\sin\sin\dfrac A2\ne0$

$\implies\cos\dfrac{B-C}2=3\sin\dfrac A2=3\cos\dfrac{B+C}2$

$\endgroup$
1
$\begingroup$

In the standard notation we obtain: $$\sin\beta+\sin\gamma=3\sin(\beta+\gamma)$$ or $$2\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}=6\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}$$ or $$\cos\frac{\beta}{2}\cos\frac{\gamma}{2}+\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=3\left(\cos\frac{\beta}{2}\cos\frac{\gamma}{2}-\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\right)$$ or $$\cos\frac{\beta}{2}\cos\frac{\gamma}{2}=2\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$$ or $$\cot\frac{\beta}{2}\cot\frac{\gamma}{2}=2.$$ Also, we can use the following way: $$\cot\frac{\beta}{2}\cot\frac{\gamma}{2}=\frac{a+b+c}{b+c-a}=\frac{4a}{2a}=2.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.