If $tan\;\theta \; = \;\frac{m}{n}$,then what is \(\frac{{msec\theta

1.	If \(tan\;\theta \; = \;\frac{m}{n}\),then what is \(\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\) equal to?1). \(\frac{{m - n}}{{m + n}}\)2). \(\frac{{{n^2} - {m^2}}}{{{n^2} + {m^2}}}\)3). \(\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\)4). 1
Answer» Given, TAN θ = m/N ⇒ Hypotenuse = √(m² + n²) $(\begin{array}{l}\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{m\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{n} - n\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}{{m\frac{{\sqrt {{m^2} + {n^2}} }}{n} + n\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}\; = \;\frac{{\frac{m}{n} - \frac{n}{m}}}{{\frac{m}{n} + \frac{n}{m}}}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\\\therefore \;\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\end{array})$

If $tan\;\theta \; = \;\frac{m}{n}$,then what is $\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}$ equal to?1). $\frac{{m - n}}{{m + n}}$2). $\frac{{{n^2} - {m^2}}}{{{n^2} + {m^2}}}$3). $\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}$4). 1

Answer»

Given,

TAN θ = m/N

⇒ Hypotenuse = √(m² + n²)

$(\begin{array}{l}\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{m\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{n} - n\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}{{m\frac{{\sqrt {{m^2} + {n^2}} }}{n} + n\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}\; = \;\frac{{\frac{m}{n} - \frac{n}{m}}}{{\frac{m}{n} + \frac{n}{m}}}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\\\therefore \;\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\end{array})$

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