Question

If tanA =ntanB and sinA= msinB prove that cos2A=m2-1/n2-1

Answer 1

tanA=ntanB
or, 1/tanB=n/tanA
or, cotB=n/tanA -----------(1)
sinA=msinB
or, 1/sinB=m/sinA
or, cosecB=m/sinA -------(2)
Now, cosec²B-cot²B=1
or, m²/sin²A-n²/tan²A=1
or, m²/sin²A-n²/(sin²A/cos²A)=1
or, m²/sin²A-n²cos²A/sin²A=1
or, (m²-n²cos²A)/sin²A=1
or, m²-n²cos²A=sin²A
or, m²-n²cos²A=1-cos²A  [∵, sin²A+cos²A=1]
or, -n²cos²A+cos²A=1-m²
or, -cos²A(n²-1)=-(m²-1)
or, cos²A=m²-1/n²-1 (Proved)

Answer 2

Answer:

Step-by-step explanation:

tanA=ntanB

or, 1/tanB=n/tanA

or, cotB=n/tanA -----------(1)

sinA=msinB

or, 1/sinB=m/sinA

or, cosecB=m/sinA -------(2)

Now, cosec²B-cot²B=1

or, m²/sin²A-n²/tan²A=1

or, m²/sin²A-n²/(sin²A/cos²A)=1

or, m²/sin²A-n²cos²A/sin²A=1

or, (m²-n²cos²A)/sin²A=1

or, m²-n²cos²A=sin²A

or, m²-n²cos²A=1-cos²A (sin²A+cos²A=1)

or, -n²cos²A+cos²A=1-m²

or, -cos²A(n²-1)=-(m²-1)

or, cos²A=m²-1/n²-1 (Proved)