0
$\begingroup$

IF $\sin \alpha = 3 \sin (\alpha+2\beta)$, then the value of $\tan (\alpha+\beta)+2 \tan \beta=$?

ATTEMPT: $\sin \alpha = 3 (\sin (\alpha+\beta) \cos \beta + \cos (\alpha+\beta) \sin \beta)$

Dividing by $\cos\beta \cos(\alpha+\beta)$ $$ \frac{\sin \alpha}{\cos \beta \cos(\alpha+\beta)}=3[\tan(\alpha+\beta)+\tan\beta] $$ Putting $\tan(\alpha+\beta)+2\tan \beta= X$ $$ \frac{\sin\alpha}{\cos\beta \cos(\alpha+\beta)}=3[X-\tan\beta] $$

$\endgroup$
0
$\begingroup$

$$SinA=3\left(SinA\,Cos2B+CosA\,Sin2B\right)$$ Divide throughout by $CosA$ we get

$$TanA=3\left(TanA\,Cos2B+Sin2B\right)$$ Use $Cos2B$ and $Sin2B$ in terms of $Tan$ and then Simplify

$\endgroup$
0
$\begingroup$

$$\frac{\sin(\alpha+2\beta)}{\sin\alpha}=\frac13$$

Applying Componendo and dividendo, $$\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}=\frac{1-3}{1+3}$$

Applying Prosthaphaeresis Formulas,

$$\frac{2\sin\beta\cos(\alpha+\beta)}{2\cos\beta\sin(\alpha+\beta)}=-2$$

Can you take it home from here?

$\endgroup$

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.