Question

If tanA=ntanB and sinA = msinB, then prove that cos^2A= m^-1 / n^+1

Answer 1

Step-by-step explanation:

tanA=ntanB

1/tanB=n/tanA

cotB=n/tanA........(1)

sinA=msinB

1/sinB=m/sinA

cosecB=m/sinA.......(2)

Now, cosec²B-cot²B=1

m²/sin²A-n²/tan²A=1

m²/sin²A-n²/(sin²A/cos²A)=1

m²/sin²A-n²cos²A/sin²A=1

(m²-n²cos²A)/sin²A=1

m²-n²cos²A=sin²A

m²-n²cos²A=1-cos²A

-n²cos²A+cos²A=1-m²

cos²A(1-n²)=1-m²

cos²A=1-m²/1-n²

cos²A=-(m²-1)/-(n²-1)

cos²A=m²-1/n²-1