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))#