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This is an extension of a question that I managed to solve but couldn’t solve the other parts.

Question:

If $\dfrac {\sin(A-B)}{\sin(A+B)}=\dfrac {5}{7}$, prove that $\dfrac{\tan(A)}{\tan(B)}=6$. If $A+B=90^\circ$, then show that $\tan(A)=\sqrt {6}$ and find the value of $\sin(2A)$. Do not use the calculator for this question.

Attempt:

I managed to prove the first part:

$$ \frac{\sin(A-B)}{\sin(A+B)}=\frac{\sin(A)\cos(B)-\sin(B)\cos(A)}{\sin(A)\cos(B)+\sin(B)\cos(A)}=\frac{5}{7} $$ Cross multiplying, $$ 7\left(\sin(A)\cos(B)-\sin(B)\cos(A)\right)= 5\left(\sin(A)\cos(B)+\sin(B)\cos(A)\right) $$ that yields $$ 2\sin(A)\cos(B)=12\sin(B)\cos(A) $$ and thus $$\frac{\tan(A)}{\tan(B)}=6$$

But I’m stuck proving $\tan(A)=\sqrt{6}$ and finding $\sin(2A)$.

Please help. Thank you!

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  • $\begingroup$ Hmmm? Did you forget the condition that $A+B=90^\circ$ for this part (the second half) of the problem? You did the difficult part (the first half) of the problem, and got stuck at the easier part. $\endgroup$
    – Batominovski
    Jul 23, 2020 at 19:34
  • $\begingroup$ $\sin(90)=1$ but I’m still stuck. Any further hint? $\endgroup$
    – IM_LOST
    Jul 23, 2020 at 19:38
  • $\begingroup$ I see. You should know that $\tan(90^\circ - \theta)=\cot(\theta)=\dfrac{1}{\tan(\theta)}$ for any angle $\theta$. See Bernard's answer below. $\endgroup$
    – Batominovski
    Jul 23, 2020 at 19:39
  • $\begingroup$ So $\tan(A+B-B)=\frac{1}{\tan(B)}$? But how does this lead to proving that $\tan(A)=\sqrt{6}$? $\endgroup$
    – IM_LOST
    Jul 23, 2020 at 19:43
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    $\begingroup$ Replace $\tan B$ with $\dfrac 1{\tan A}$ in the ratio. $\endgroup$
    – Bernard
    Jul 23, 2020 at 19:57

1 Answer 1

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Hint

  1. Complementary angles have tangents inverse of each other.
  2. Use the duplication formula: $$\sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta}.$$
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