Question

If $tan \theta + sin \theta = m$ and $tan \theta - sin \theta = n$, then show that $(m² - n²)² = 16mn \: or \: (m² - n²) = 4 \sqrt{mn}$.​

Answer 1

GIVEN :-

$\\ \bullet \: \: \: \: \: \: \: \sf \: tan \theta+ sin\theta =m \\ \bullet \: \: \: \: \: \: \: \sf \: tan\theta - sin\theta = n$

TO KNOW :-

• sin²A + cos²A = 1
• sinA/cosA = tanA
• (x+y)² = x² + y² + 2xy
• (x-y)² = x² + y² - 2xy

TO PROVE :-

$\\ \sf \: \{ {m}^{2} - { {n}^{2} \}}^{2} = 16mn$

SOLUTION :-

$\\ \sf \: L.H.S = \{ {m}^{2} - { {n}^{2} \}}^{2}$

Substituting values of m and n ,

$\\ \\ \implies \sf \{ {(tan\theta + sin\theta)}^{2} - ( \ {tan\theta - sin\theta) ^{2} \}}^{2} \\ \\ \implies \sf \: \{ ({tan}^{2}\theta + {sin}^{2} \theta + 2tan\theta.sin\theta) - ( {tan}^{2} \theta + {sin}^{2}\theta - 2tan\theta.sin\theta) \} ^{2} \\ \\ \implies \sf \: \{ \cancel{{tan}^{2}\theta} + \cancel{ {sin}^{2}\theta} + 2tan\theta.sin\theta - \cancel{{tan}^{2}\theta } - \cancel{ {sin}^{2}\theta } + 2tan\theta.sin\theta \} ^{2} \\ \\ \implies \sf \: \{4tan\theta.sin\theta \} ^{2} \\ \\ \implies \sf \: 16 {tan}^{2}\theta . {sin}^{2} \theta \: \: \: \: \: \: \: - - - (1)$

Now we will solve R.H.S

$\\ \sf \: R.H.S = 16mn$

Substituting values of m and n ,

$\\ \implies \sf \: 16(tan\theta + sin\theta)(tan\theta - sin\theta) \\ \\ \\ \implies \sf \: 16( {tan}^{2} \theta - {sin}^{2}\theta ) \\ \\ \\ \implies \sf \: 16 \left( \frac{ {sin}^{2}\theta }{ {cos}^{2}\theta } - {sin}^{2} \theta \right ) \\ \\ \\ \implies \sf \: 16 \left( \frac{ {sin}^{2} \theta - {sin}^{2}\theta . {cos}^{2}\theta }{ {cos}^{2}\theta } \right) \\ \\ \\ \implies \sf \: 16 \left( \frac{ {sin}^{2}\theta (1 - {cos}^{2}\theta )}{ {cos}^{2} \theta} \right) \\ \\ \\ \implies \sf \: 16 \left( \frac{ {sin}^{2}\theta . {sin}^{2}\theta }{ {cos}^{2}\theta } \right)$

$\\ \\ \implies \sf \: 16 {sin}^{2}\theta. {tan}^{2}\theta \: \: \: \: \: \: \: \: - - - (2)$

By equation (1) and (2) ,

LHS = RHS

$\\ \boxed{ \sf\{ {m}^{2} - {n}^{2} \} ^{2} = 16mn}$

Taking root ,

$\\ \\ \boxed{ \sf \: {m}^{2} - {n}^{2} = 4 \sqrt{mn} }$ (proved)