1. tan and cos

if (a+b)tan(theta-phi) = (a-b)tan(theta + phi) and a cos2 phi + b cos2 theta =c then prove that a square - b square + c square = 2 ac cos2 phi.

my soln.
a+b/a-b = tan(t + p)/tan(t-p)

2. Re: tan and cos

Rewrite:
$\displaystyle a^2-b^2+c^2=2ac\cos(2\phi)$ as
$\displaystyle \left[(a-b)(a+b)\right]+c^2=2ac\cos(2\phi)$

Maybe you can do something with this.
...

3. Re: tan and cos

Originally Posted by Siron
Rewrite:
$\displaystyle a^2-b^2+c^2=2ac\cos(2\phi)$ as
$\displaystyle \left[(a-b)(a+b)\right]+c^2=2ac\cos(2\phi)$

Maybe you can do something with this.
...
not being able to proceed.

4. Re: tan and cos

Originally Posted by saha.subham
if $\displaystyle (a+b)\tan(\theta-\phi) = (a-b)\tan(\theta + \phi)$ and $\displaystyle a \cos(2 \phi) + b \cos(2 \theta) =c$ then prove that $\displaystyle a^2-b^2+c^2 = 2ac\cos(2\phi).$
In the first equation, write the tans as sin/cos and then use the trig formulas for sums and products, to get

$\displaystyle (a+b)\sin(\theta-\phi)\cos(\theta + \phi) = (a-b)\sin(\theta+\phi)\cos(\theta - \phi),$

$\displaystyle (a+b)\bigl(\sin(2\theta)-\sin(2\phi)\bigr) = (a-b)\bigl(\sin(2\theta) + \sin(2\phi)\bigr),$

$\displaystyle b\sin(2\theta) = a\sin(2\phi).\qquad(1)$

You also have the equation $\displaystyle a \cos(2 \phi) + b \cos(2 \theta) =c$, from which

$\displaystyle b \cos(2 \theta) = c - a \cos(2 \phi).\qquad(2)$

Square the equations (1) and (2) and add:

$\displaystyle b^2 = a^2\sin^2(2\phi) + (c - a \cos(2 \phi))^2.$

That simplifies to give the result that you want.

Notice that this is really a problem about a triangle, with sides a, b, c and angles $\displaystyle 2\theta$ (opposite the side a) and $\displaystyle 2\phi$ (opposite the side b). The equation (1) is then the sine rule, and the equation $\displaystyle a^2-b^2+c^2 = 2ac\cos(2\phi)$ is the cosine rule.

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if (a b) tan(theta-phi)=(a-b)

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