Question

If (tan θ + sin θ)=m and (tan θ-sin θ)=n, prove that --
$({m}^{2} - {n}^{2})^{2} = 16mn$
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Answer 1

Answer:

Step-by-step explanation:

(tan θ + sin θ)=m & (tan θ-sin θ)=n

m²=tan² θ + sin² θ + 2 tan θ sin θ

n²=tan² θ + sin² θ - 2 tan θ sin θ

Subtracting,

m²-n²= 4 tan θ sin θ

Squaring,

(m²-n²)² = 16 tan² θ sin² θ .....(1)

Also, mn = (tan θ + sin θ)(tan θ-sin θ) =(tan² θ - sin² θ)

mn = (sin² θ/cos²θ - sin²θ) = sin²θ(1/cos²θ - 1) = sin²θ(1-cos²θ)/cos²θ

or mn = (sin²θ/cos²θ) (1-cos²θ) = tan²θ sin²θ ......(2)

From (1) & (2), we have

(m²-n²)² = 16mn

Hence proved.