Question

If tan titha+ sin titha=m tan titha-sin titha=n
Then find
M square+N square =4√mn

Answer 1

Question:

If tan θ + sin θ = m, tanθ - sin θ =n, then prove  that m² - n² = 4√mn

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• tan θ + sin θ = m

• tanθ - sin θ =n

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• m² - n² = 4√mn

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Taking the LHS of the equation,

   m² - n² = (tan θ + sin θ)² - (tan θ -sin θ)²

→ Expanding it we get,

    m² - n² = tan²θ + 2 sinθ tanθ + sin²θ - tan²θ + 2 sinθ tanθ - sin²θ

→ Cancelling the like terms we get,

     m² - n² = 4 sinθ tanθ-----equation 1

→ Now taking the RHS of the equation,

  $4\sqrt{mn}=4\times \sqrt{(tan\theta +sin\theta)(tan\theta-sin\theta)}$

  $4\sqrt{mn}=4\times \sqrt{tan^{2}\theta-sin^{2}\theta }$

  $4\sqrt{mn} =4\times \sqrt{\dfrac{sin^{2}\theta }{cos^{2}\theta }+sin^{2}\theta }$

  $4\sqrt{mn}=4\times \sqrt{sin^{2}\theta (\dfrac{1}{cos^{2}\theta }-1) }$

  $4\sqrt{mn} = 4sin\theta\sqrt{sec^{2}\theta-1 }$

  $4\sqrt{mn} =4sin\theta \: tan\theta-equation\:2$

→ From equation 1 and 2, RHS are equal, therefore LHS must also be equal

  m² - n² = 4√mn

Hence proved.

$\Large{\underline{\underline{\bf{Identities\:used:}}}}$

→ ( a + b )² = a² + 2ab + b²

→ ( a - b )² = a² - 2ab + b²

→ a² - b² = ( a + b ) × ( a - b )

→ tan θ = sin θ/ cos θ

→ sec θ = 1/ cos θ

→ sec²θ - 1 = tan²θ