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

Question

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

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Answer 1 · 2017-08-13T01:29:24

Given

$sina/sinb=p and cosa/cosb=q$

We have

$sina=p*sinb ........[1]$

and

$cosa=q*cosb.....[2]$

So
$sin^2a+cos^2a=1$

$=>p^2sin^2b+q^2cos^2b=1$

$=>p^2tan^2b+q^2=sec^2b$

$=>p^2tan^2b+q^2=1+tan^2b$

$=>(p^2-1)tan^2b=(1-q^2)$

$=>tan^2b=(1-q^2)/(p^2-1)$

$=>tanb=pmsqrt((1-q^2)/(p^2-1))......[3]$

Similarly we can proceed for $tana$

Given

$sina/sinb=p and cosa/cosb=q$

We have

$sinb=sina/pand cosb=cosa/q$

So
$sin^2b+cos^2b=1$

$=>sin^2a/p^2+cos^2a/q^2=sin^2a+cos^2a$

$=>sin^2a(1/p^2-1)=cos^2a(1-1/q^2)$

$=>sin^2a/cos^2a=(1-1/q^2)/(1/p^2-1)$

$=>tan^2a=(1-1/q^2)/(1/p^2-1)$

$=>tana=pmsqrt((1-1/q^2)/(1/p^2-1))=pmp/qsqrt((q^2-1)/(1-p^2))$

$=>tana=pmp/qsqrt((1-q^2)/(p^2-1))$

Otherwise

Dividing [1] by [2]we have

$tana=p/qtanb$

using {3] we get

$tana=pmp/qsqrt((1-q^2)/(p^2-1))$