Question

$2mn \: sin \: theta = ({m}^{2} - {n}^{2} ) \cos \: theta \: then \: an \: incorrect \: statement \: is \: \\ a) \sec \: thete = \frac{ {m - {n}^{2} }^{2} }{2mn } \\ \\ b) \tan \: theta = \frac{ {m + n}^{2} }{ {m - n }^{2} } \\ c) csc \: theta = \frac{ {m - n}^{2} }{ {m + n}^{2} } \\ d) \sin \: theta \frac{ {m - n}^{2} }{ {m + n2}^{2} } \\ please \: answer \: fast$​

Answer 1

Given :

$\bullet\ \; \sf 2mn.sin \theta=(m^2-n^2)cos \theta$

To Find :

Wrong option given

Solution :

Note :

P  ↔  Perpendicular

B  ↔  Base

H  ↔  Hypotenuse

_________________

$\sf 2mn.sin \theta=(m^2-n^2)cos \theta\\\\\to \sf \dfrac{sin\theta}{cos\theta}=\dfrac{m^2-n^2}{2mn}\\\\\to \sf tan\theta=\dfrac{m^2-n^2}{2mn}$

Here ,

• P = m² - n²  
• B = 2mn

Apply Pythagoras theorem ,

⇒ H² = P² + B²

⇒ H² = ( m² - n² )² + ( 2mn )²

⇒ H² = m⁴ + n⁴ - 2m²n² + 4m²n²

⇒ H² = (m²)² + (n²)² + 2(m²)(n²)

⇒ H² = ( m² + n² )²

⇒ H = m² + n²

P = m² - n²  

B = 2mn

________________________

$\sf sec\theta=\dfrac{H}{B}\\\\\to \sf sec\theta=\dfrac{m^2+n^2}{2mn}$

$\sf tan\theta=\dfrac{P}{B}\\\\\to \sf tan\theta=\dfrac{m^2-n^2}{2mn}$

$\sf csc\theta=\dfrac{H}{P}\\\\\to \sf csc\theta=\dfrac{m^2+n^2}{m^2-n^2}$

$\sf sin\theta=\dfrac{P}{H}\\\\\to \sf sin\theta=\dfrac{m^2-n^2}{m^2+n^2}$

Option b) is wrong here

Answer 2

Given :

$\bullet\ \; \sf 2mn.sin \theta=(m^2-n^2)cos \theta$

To Find :

Wrong option given

Solution :

$\begin{gathered}\sf 2mn.sin \theta=(m^2-n^2)cos \theta\\\\\to \sf \dfrac{sin\theta}{cos\theta}=\dfrac{m^2-n^2}{2mn}\\\\\to \sf tan\theta=\dfrac{m^2-n^2}{2mn}\end{gathered}$

Here ,

P = m² - n²

B = 2mn

Apply Pythagoras theorem ,

⇒ H² = P² + B²

⇒ H² = ( m² - n² )² + ( 2mn )²

⇒ H² = m⁴ + n⁴ - 2m²n² + 4m²n²

⇒ H² = (m²)² + (n²)² + 2(m²)(n²)

⇒ H² = ( m² + n² )²

⇒ H = m² + n²

P = m² - n²

B = 2mn

$\begin{gathered}\sf sec\theta=\dfrac{H}{B}\\\\\to \sf sec\theta=\dfrac{m^2+n^2}{2mn}\end{gathered}$

$\begin{gathered}\sf tan\theta=\dfrac{P}{B}\\\\\to \sf tan\theta=\dfrac{m^2-n^2}{2mn}\end{gathered}$

$\begin{gathered}\sf csc\theta=\dfrac{H}{P}\\\\\to \sf csc\theta=\dfrac{m^2+n^2}{m^2-n^2}\end{gathered}$

$\begin{gathered}\sf sin\theta=\dfrac{P}{H}\\\\\to \sf sin\theta=\dfrac{m^2-n^2}{m^2+n^2}\end{gathered}$

Option b) is incorrect here

NOTE

P ↔ Perpendicular

B ↔ Base

H ↔ Hypotenuse