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I have the following identity:

$$ \frac{\tan (t + h) - \tan(t)}{h} = \left( \frac{\tan (h)}{h} \right)\left( \frac{\sec^2(t)}{1 - \tan (t)\tan (h)} \right)$$


Having tried various approaches, which are far too varied and numerous to list all here, the only one that seems to have the most promise is this one (however it still falls quite short); using the right hand side:

$$\begin{align}\text{LHS} &= \left( \frac{\tan (h)}{h} \right)\left( \frac{\sec^2(t)}{1 - \tan (t)\tan (h)} \right)\\ &=\frac{\dfrac{\sin (h)}{\cos(h)}}{h} \cdot \dfrac{\dfrac{1}{\cos^2(t)}}{\dfrac{\cos(t)\cos(h)-\sin(t)\sin(h)}{\cos(t)\cos(h)}}\\ &= \dfrac{1}{h} \cdot \dfrac{\dfrac{\sin(h)}{\cos(h)\cos^2(t)}}{\dfrac{\cos(t)\cos(h)-\sin(t)\sin(h)}{\cos(t)\cos(h)}}\\ &= \frac{1}{h} \cdot \frac{\sin (h)}{\cos(h)\cos ^2(t)}\cdot \frac{\cos (t)\cos(h)}{\cos(t)\cos(h)-\sin(t)\sin(h)}\\ &= \frac{\sin(h)}{h\cdot\cos(t)\cdot\cos(t+h)}\end{align}$$

Can anyone throw me a bone here? I'm stumped. Thanks.

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    $\begingroup$ Really, don't use \huge. It's terrible. $\endgroup$ – Thomas Andrews Jul 27 '14 at 15:48
  • $\begingroup$ If you can clean it up please do so, due to all the fractions the text had become quite small so I was worried no one would be able to read it. $\endgroup$ – seeker Jul 27 '14 at 15:50
  • $\begingroup$ @Assad Try using \$\displaystyle{...}\$ instead of \$\big\$ or \$\huge\$ etc. if an excess of fractions causes your maths to be unreadable. $\endgroup$ – Jam Jul 27 '14 at 16:11
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Apply the rule

$$ \tan(t+h) = \frac{\tan(t) + \tan(h)}{1-\tan(t)\tan(h)} $$

to the LHS and it's straightforward from there.

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  • $\begingroup$ Is it not possible to use the right hand side? $\endgroup$ – seeker Jul 27 '14 at 15:52
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\begin{align} LHS &=\frac{\tan(t+h)-\tan t}{h}\\ &=\frac{1}{h}\left(\frac{\tan{t}+\tan h}{1-\tan t \tan h}-\frac{\tan t-\tan^2 t\tan h}{1-\tan t \tan h}\right)\\ &=\frac{1}{h}\frac{\tan{h}(1+\tan^2 t)}{1-\tan t \tan h}\\ &=\frac{\tan h}{h}\frac{\sec^2 t}{1-\tan t \tan h}\\ &=RHS \end{align}

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$$\frac{\tan(x+h)-\tan x}h=\frac{\sin(x+h-x)}{h\cos x\cos(x+h)}=\frac{\tan h}h\frac{\cos h}{\cos x\cos(x+h)}$$ Now, $$\frac{\cos h}{\cos x\cos(x+h)}=\frac{\cos h}{\cos x(\cos x\cos h-\sin x\sin h)} =\frac{\cos h}{\cos^2x\cos h(1-\tan x\tan h)}$$

Multiply the numerator & the denominator by $\sec^2x$

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  • $\begingroup$ How did the numerator go from $\tan(x+h)-\tan (x)$ to $\sin (x+h-x) $? $\endgroup$ – seeker Jul 27 '14 at 18:51
  • $\begingroup$ @Assad, Write $$\tan =\frac{\sin }{\cos }$$ $\endgroup$ – lab bhattacharjee Jul 28 '14 at 3:10

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