## shun 3 years ago how can we derive thew formula for half angle for tangent.. pls help..

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1. IsTim

2. IsTim

The derivaive of TAN is cosx/sinx I think? Or is that just the reciprocal?

3. eyust707

$\tan \theta = {\sin \theta \over \cos \theta}$

4. eyust707

The derivative of tangent is: $\frac{ d }{ d \theta } \tan \theta = \sec^2\theta$

5. qpHalcy0n

I think he's looking for a proof of the tangent half angle formula, not an actual derivative.

6. ash2326

@shun you want to derive $$\tan \frac {\theta}{2}$$??

7. ash2326

We know that $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\times \tan B}$ Put $A=B=\frac {\theta}{2}$ $$\large \text{you'd get a relation of}\ \tan \theta \ \text{in terms of} \tan \frac{\theta}2$$

8. qpHalcy0n

Here, to prove it, use the half angle formulae for sine and cosine. $\sin(x/2) = \pm \sqrt{ (1-cosx / 2}$ $\cos(x/2) = \pm \sqrt{ (1+cosx / 2) }$

9. shun

@ash yes,, actually,, it is,

10. ash2326

you got it?

11. shun

yes i got it,, tnx,,

12. satellite73

$\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}$

13. shun

@qp tnx.. i get it.. tank you.. but that it is the formulaim asking for the derivation.. ..

14. qpHalcy0n

The derivation is based on those formulae. The half angle tangent is half angle sine / half angle cosine. You can prove half angle sine and cosine as well, if needs be.

15. shun

@sat.. yes that the beginning.. did you know some restrictions?

16. shun

@qp ok ok.. i get it.. tnx,,,