If tan A=n tan B and sin A = m sin B then show that cos²A=(m²-1)/(n²-1) ?

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Answer 1 · 2017-12-02T17:53:07

Kindly refer to a Proof in the Explanation.

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

#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.!