Question

PLZZ SOLVE THIS::-
IF TAN A = n TAN B AND
Sin A = m Sin B
PROVE THAT :::::
COSA ^2 = m ^2 - 1 / n ^2 - 1 ​

Answer 1

Given that, tanA=ntanB rArr sinA/cosA=n*sinB/cosB.

tanA=ntanB⇒sinAcosA=n⋅sinBcosB.

rArr sinA/sinB=n*cosA/cosB.....................................<<1>>.

Also given that, sinA=msinB rArr sinA/sinB=m.....................<<2>>.

Comparing <<1>> and <<2>>, we get,

m=n*cosA/cosB," giving, "cosB=n/m*cosA......<<3>>.

<<2>> rArr sinB=1/m*sinA...................................<<2'>>.

Now, using <<2'>> and <<3>> in cos^2B+sin^2B=1, we get,

rArrn^2/m^2*cos^2A+1/m^2*sin^2A=1.

rArr n^2cos^2A+sin^2A=m^2.

rArr n^2cos^2A+(1-cos^2A)=m^2.

rArr n^2cos^2A-cos^2A=m^2-1, i.e.,

rArr (n^2-1)cos^2A=(m^2-1).

rArr cos^2A=(m^2-1)/(n^2-1),

as desired!

Enjoy Maths.!