# If sin(a)/sin(b) = p and cos(a)/cos(b) = q then tan(b) =?

## What about tan(a)? That shd also be expressed in p and q in the answer

##### 1 Answer
Aug 13, 2017

Given

$\sin \frac{a}{\sin} b = p \mathmr{and} \cos \frac{a}{\cos} b = q$

We have

$\sin a = p \cdot \sin b \ldots \ldots . . \left[1\right]$

and

$\cos a = q \cdot \cos b \ldots . . \left[2\right]$

So
${\sin}^{2} a + {\cos}^{2} a = 1$

$\implies {p}^{2} {\sin}^{2} b + {q}^{2} {\cos}^{2} b = 1$

$\implies {p}^{2} {\tan}^{2} b + {q}^{2} = {\sec}^{2} b$

$\implies {p}^{2} {\tan}^{2} b + {q}^{2} = 1 + {\tan}^{2} b$

$\implies \left({p}^{2} - 1\right) {\tan}^{2} b = \left(1 - {q}^{2}\right)$

$\implies {\tan}^{2} b = \frac{1 - {q}^{2}}{{p}^{2} - 1}$

$\implies \tan b = \pm \sqrt{\frac{1 - {q}^{2}}{{p}^{2} - 1}} \ldots \ldots \left[3\right]$

Similarly we can proceed for $\tan a$

Given

$\sin \frac{a}{\sin} b = p \mathmr{and} \cos \frac{a}{\cos} b = q$

We have

$\sin b = \sin \frac{a}{p} \mathmr{and} \cos b = \cos \frac{a}{q}$

So
${\sin}^{2} b + {\cos}^{2} b = 1$

$\implies {\sin}^{2} \frac{a}{p} ^ 2 + {\cos}^{2} \frac{a}{q} ^ 2 = {\sin}^{2} a + {\cos}^{2} a$

$\implies {\sin}^{2} a \left(\frac{1}{p} ^ 2 - 1\right) = {\cos}^{2} a \left(1 - \frac{1}{q} ^ 2\right)$

$\implies {\sin}^{2} \frac{a}{\cos} ^ 2 a = \frac{1 - \frac{1}{q} ^ 2}{\frac{1}{p} ^ 2 - 1}$

$\implies {\tan}^{2} a = \frac{1 - \frac{1}{q} ^ 2}{\frac{1}{p} ^ 2 - 1}$

$\implies \tan a = \pm \sqrt{\frac{1 - \frac{1}{q} ^ 2}{\frac{1}{p} ^ 2 - 1}} = \pm \frac{p}{q} \sqrt{\frac{{q}^{2} - 1}{1 - {p}^{2}}}$

$\implies \tan a = \pm \frac{p}{q} \sqrt{\frac{1 - {q}^{2}}{{p}^{2} - 1}}$

Otherwise

Dividing  by we have

$\tan a = \frac{p}{q} \tan b$

using {3] we get

$\tan a = \pm \frac{p}{q} \sqrt{\frac{1 - {q}^{2}}{{p}^{2} - 1}}$