# Proving $\lim_{\beta\to\alpha}\frac{\alpha\sin\beta-\beta\sin\alpha}{\alpha \cos \beta- \beta \cos \alpha}= \tan(\alpha-\tan^{-1}\alpha)$ [closed]

Prove that $$\lim_{\beta\to\alpha}\;\frac{\alpha\sin\beta-\beta\sin\alpha}{\alpha \cos \beta- \beta \cos \alpha}= \tan(\alpha-\tan^{-1}\alpha)$$

I am solving the exercise from the S.L. Loney plane trigonometry book, page 48 question number 35, I got stuck.

Any help will be truly appreciated.

• L’Hôpital’s rule? May 4 at 19:21
• I used series expansion and tried to do it. We are not allowed to use derivatives. May 4 at 19:27
• Do you mean the limit as $\beta$ tends to $a$? May 4 at 19:35
• When you got the limit, you can derive the required formula using $$tan(x-y)= \frac{tanx-tany}{1+tanx \cdot tany}$$ May 4 at 20:01

Hint:

Write the numerator as

$$a(\sin \beta-\sin a)-(\beta-a)\sin a=(\beta-a)\left(a\cdot\dfrac{\sin\beta-\sin a}{\beta -a}-\sin a\right)$$

Similarly for the denominator

Alternatively, replace $$\beta$$ with $$a+h$$ to find the numerator $$=a\sin(a+h)-(a+h)\sin a=a(\sin(a+h)-\sin a)-h\sin a$$

Similarly for the denominator

Now divide the numerator and the denominator by $$h$$ as $$h\ne0$$ as $$h\to0$$

Considering $$\frac{\alpha\sin(\beta)-\beta\sin(\alpha)}{\alpha \cos (\beta)- \beta \cos (\alpha)}$$ make life simpler letting $$\beta=\alpha+x$$ to face the problem of the limit, when $$x\to 0$$ of $$\frac{\sin (\alpha ) (\alpha +x)-\alpha \sin (\alpha +x)}{\cos (\alpha ) (\alpha +x)-\alpha \cos (\alpha +x)}$$ Expanding, numerator and denominator $$\text{num}=\alpha \sin (\alpha )+x \sin (\alpha )-\alpha \cos (\alpha ) \sin (x)-\alpha \sin (\alpha ) \cos (x)$$ $$\text{den}=\alpha \cos (\alpha )+\alpha \sin (\alpha ) \sin (x)+x \cos (\alpha )-\alpha \cos (\alpha ) \cos (x)$$

Now, use Taylor series or, simpler, equivalents to make $$\frac{\text{num}}{\text{den}}\sim \frac {x (\sin (\alpha )-\alpha \cos (\alpha )) } {x (\alpha \sin (\alpha )+\cos (\alpha )) }=\frac{\sin (\alpha )-\alpha \cos (\alpha )}{\alpha \sin (\alpha )+\cos (\alpha )}=\frac{ \tan (\alpha )-\alpha}{1+\alpha \tan (\alpha ) }$$ and recognize a well known trigonometric formula.