Question

solve this
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Answer 1

tanA=ntanB⇒sinAcosA=n⋅sinBcosB.

⇒sinAsinB=n⋅cosAcosB.....................................⟨1⟩.

Also given that, sinA=msinB⇒sinAsinB=m.....................⟨2⟩.

Comparing ⟨1⟩and⟨2⟩, we get,

m=n⋅cosAcosB, giving, cosB=nm⋅cosA......⟨3⟩.

⟨2⟩⇒sinB=1m⋅sinA...................................⟨2'⟩.

Now, using ⟨2'⟩and⟨3⟩ in cos2B+sin2B=1, we get,

⇒n2m2⋅cos2A+1m2⋅sin2A=1.

⇒n2cos2A+sin2A=m2.

⇒n2cos2A+(1−cos2A)=m2.

⇒n2cos2A−cos2A=m2−1,i.e.,

⇒(n2−1)cos2A=(m2−1).

⇒cos2A=m2−1n2−1, as desired!

Enjoy Maths.!