[tex] 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[/tex]
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Given :
[tex]\bullet\ \; \sf 2mn.sin \theta=(m^2-n^2)cos \theta[/tex]
To Find :
Wrong option given
Solution :
Note :
P ↔ Perpendicular
B ↔ Base
H ↔ Hypotenuse
_________________
[tex]\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}[/tex]
Here ,
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
________________________
[tex]\sf sec\theta=\dfrac{H}{B}\\\\\to \sf sec\theta=\dfrac{m^2+n^2}{2mn}[/tex]
[tex]\sf tan\theta=\dfrac{P}{B}\\\\\to \sf tan\theta=\dfrac{m^2-n^2}{2mn}[/tex]
[tex]\sf csc\theta=\dfrac{H}{P}\\\\\to \sf csc\theta=\dfrac{m^2+n^2}{m^2-n^2}[/tex]
[tex]\sf sin\theta=\dfrac{P}{H}\\\\\to \sf sin\theta=\dfrac{m^2-n^2}{m^2+n^2}[/tex]
Option b) is wrong here
Given :
[tex]\bullet\ \; \sf 2mn.sin \theta=(m^2-n^2)cos \theta[/tex]
To Find :
Wrong option given
Solution :
[tex]\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}[/tex]
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
[tex]\begin{gathered}\sf sec\theta=\dfrac{H}{B}\\\\\to \sf sec\theta=\dfrac{m^2+n^2}{2mn}\end{gathered}[/tex]
[tex]\begin{gathered}\sf tan\theta=\dfrac{P}{B}\\\\\to \sf tan\theta=\dfrac{m^2-n^2}{2mn}\end{gathered}[/tex]
[tex]\begin{gathered}\sf csc\theta=\dfrac{H}{P}\\\\\to \sf csc\theta=\dfrac{m^2+n^2}{m^2-n^2}\end{gathered}[/tex]
[tex]\begin{gathered}\sf sin\theta=\dfrac{P}{H}\\\\\to \sf sin\theta=\dfrac{m^2-n^2}{m^2+n^2}\end{gathered} [/tex]
Option b) is incorrect here
NOTE
P ↔ Perpendicular
B ↔ Base
H ↔ Hypotenuse