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If + = m and - = n , show that m^2-n^2 = 4 { mn}

Question

If $tan θ + sin θ = m$ and $tan θ - sin θ = n$ , show that

$m^{2} - n^{2} = 4 mn$

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Harper Discussed

tan θ + sin θ = m

and

tan θ - sin θ = n

, show that

m^{2} - n^{2} = 4 mn

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Given,

tan θ + sin θ = m

and

tan θ - sin θ = n

we have,

m + n = 2 tan θ

m - n = 2 sin θ

m^{2} - n^{2} = (m + n) (m - n)

\Rightarrow m^{2} - n^{2} = 4 sin θ tan θ

..........(1)

mn = (tan θ + sin θ) (tan θ - sin θ)

\Rightarrow mn = tan^{2} θ - sin^{2} θ

now,

4 mn = 4 tan^{2} θ - sin^{2} θ

= 4 \frac{sin ^{2} θ}{cos ^{2} θ} - sin^{2} θ

\Rightarrow 4 mn = 4 sin θ tan θ

..........(2)

equating (1) and (2), we get,

m^{2} - n^{2} = 4 mn

Hence proved.

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Question Text	If $tan θ + sin θ = m$ and $tan θ - sin θ = n$ , show that $m^{2} - n^{2} = 4 mn$
Answer Type	Text solution:1
Upvotes	150