Vertex $A$ of $\triangle ABC$ moves in such a way that $\tan B+\tan C = c(\bf{constant}),$
Then locus of orthocentre of $\triangle ABC$ is ( side $BC$ is fixed)
$\bf{My\; Try::}$ We can write $$\tan B+\tan C = c\Rightarrow \frac{\sin B}{\cos C}+\frac{\sin C}{\cos B} = c$$
So $$\frac{\sin (B+C)}{\cos B \cos C} = c\Rightarrow \sin (B+C)=c \cdot \cos B \cos C$$
Using $A+B+C = \pi\Rightarrow A+B = \pi-C$
So $$\sin (\pi-A) = c\cdot \cos B \cos C\Rightarrow \sin A = c\cdot \cos B\cdot \cos C$$
Now how can i solve it after that, Help required, Thanks