Question

$if \\ \tan( \theta) + \sin(\theta) = m \\\tan(\theta)+\sin(\theta) = n \\ \\ show \: that \: {m}^{2} - {n}^{2} = 4 \sqrt{mn}$

Answer 1

hey, ur question is wrong but here is ur ans. of correct question.....

Answer 2

Answer:

m² - n² = 4√mn

Step-by-step explanation:

Note: Your questions seems to be incorrect.It should be tanθ-sinθ=n.

Given Equations are:

(i) tanθ + sinθ = m.

(ii) tanθ - sinθ = n  

(iii)

On adding (i) & (ii), we get

⇒ tanθ + sinθ + tanθ - sinθ = m + n

⇒ 2tanθ = m + n.

(iv)  

On subtracting (i) & (ii), we get

⇒ tanθ + sinθ - tanθ + sinθ = m - n

⇒ 2sinθ = m - n.

On solving (iii) & (iv), we get

2 tanθ = m + n

2 sinθ = m - n

-----------------------

4 tanθ sinθ = (m + n)(m - n)

4 tanθ sinθ = m² - n²

4√mn = m² - n²

4√tan²θsin²θ = m² - n²

4√(sin²θ/cos²θ) * sin²θ = m² - n²

4√(1 - cos²θ/cos²θ) * sin²θ = m² - n²

4√(sin²θ - sin²θcos²θ/cos²θ) = m² - n²

4√(sin²θ/cos²θ) - sin²θ = m² - n²

4√tan²θ - sin²θ = m² - n²

4√(tanθ + sinθ)(tanθ - sinθ) = m² - n²

4√mn = m² - n²

Hope it helps!