Question

If tanA +sinA =m and tanA -sinA=n show that (m^2-n^2) =\sqrt{16}mn

Answer 1

Answer:

Solution:

• Given : tanA + sinA = m -------(1)

tanA - sinA = n --------(2)

• To prove : m² - n² = √(16mn)

• Proof :

Adding eq-(1) and (2) , we get ;

=> tanA + sinA + tanA - sinA = m + n

=> 2tanA = m + n --------(3)

Now,

Subtracting eq-(2) from (1) , we get ;

=> (tanA + sinA) - (tanA - sinA) = m - n

=> tanA + sinA - tanA + sinA = m - n

=> 2sinA = m - n -----------(4)

Multiplying eq-(3) and (4) , we get ;

=> 2tanA•2sinA = (m + n)(m - n)

=> 4sinA•tanA = m² - n² -------(5)

Now,

Multiplying eq-(1) and (2) , we get ;

=> (tanA + sinA)(tanA - sinA) = mn

=> tan²A - sin²A = mn

=> sin²A/cos²A - sin²A = mn

=> sin²A(1/cos²A - 1) = mn

=> sin²A(1 - cos²A)/cis²A = mn

=> (sin²A/cos²A)(1 - cos²A) = mn

=> tan²A•sin²A = mn

=> (sinA•tanA)² = mn

=> sinA•tanA = √(mn)

Now ,

Putting sinA•tanA = √mn in eq-(5) , we get ;

=> 4sinA•tanA = m² - n²

=> 4√(mn) = m² - n²

=> √(16mn) = m² - n²

=> m² - n² = √(16mn)

Hence proved .