Question

pls answer fast ,it's very urgent...​

Answer 1

$cos^{2} x = \frac{m^{2}-1 }{n^{2}-1}$

Step-by-step explanation:

$tanx$ = n $tany$

therefore,  

n = $\frac{tanx}{tany}$ (Equation 1)

and $sinx$ = m $siny$

therefore,  

m = $\frac{sinx}{siny}$ (Equation 2)

L.H.S = $cos^{2} x </p><p>R.H.S = [tex]\frac{m^{2}-1 }{n^{2}-1}$

Taking R.H.S

\frac{m^{2}-1 }{n^{2}-1}[/tex]

Replace the value of m  & n by the Equation 1 & Equation 2 Respectively.

= $\frac{(\frac{sinx}{siny}) ^{2}- 1}{(\frac{tanx}{tany})^{2}-1 }$

= $\frac{\frac{sin^{2}x-sin^{2}y  }{sin^{2}y } }{\frac{tan^{2}x -tan^{2}y  }{tan^{2} y} }$

as we know $tan^{2} y = \frac{sin^{2}y }{cos^{2} y}$

therefore,

=    $\frac{\frac{sin^{2}x-sin^{2}y  }{sin^{2}y } }{\frac{tan^{2}x -   tan^{2}y  }{\frac{sin^{2}y}{cos^{2} y}} }$

=$\frac{sin^{2}x-sin^{2}y }{sin^{2}y }$   $\frac{sin^{2}y }{cos^{2}y(tan^{2}x-tan^{2}y)}$

=$\frac{sin^{2}x-sin^{2}y }{cos^{2}y(tan^{2}x-tan^{2}y)}$

we know that,

= $sin^{2} x$=$1-cos^{2} x$

therefore replace $sin^{2} x$=$1-cos^{2} x$

=$\frac{1-cos^{2}x-1+cos^{2}y }{cos^{2}y(tan^{2}x-tan^{2}y)}$

we know that,

=$tan^{2} x=sec^{2} x-1$

thus,

=$\frac{cos^{2}y-cos^{2}x }{cos^{2}y(sec^{2}x-1-sec^{2}y+1)}$

=$\frac{cos^{2}y-cos^{2}x }{cos^{2}y(\frac{1}{cos^{2}x } -\frac{1}{cos^{2}y})}$

=$\frac{cos^{2}y-cos^{2}x }{cos^{2}y(\frac{cos^{2}y -cos^{2}x }{cos^{2}x .cos^{2}y})}$

= $cos^{2}x$ =L.H.S

Answer 2

Answer:

cos^{2} x = \frac{m^{2}-1 }{n^{2}-1}cos

2

x=

n

2

−1

m

2

−1

Step-by-step explanation:

tanxtanx = n tanytany

therefore,

n = \frac{tanx}{tany}

tany

tanx

(Equation 1)

and sinxsinx = m sinysiny

therefore,

m = \frac{sinx}{siny}

siny

sinx

(Equation 2)

L.H.S = cos^{2} x < /p > < p > R.H.S = $\frac{m^{2}-1 }{n^{2}-1}cos </p><p>2</p><p> x</p><p>R.H.S=[tex] </p><p>n </p><p>2</p><p> −1</p><p>m </p><p>2</p><p> −1</p><p> </p><p> </p><p></p><p>Taking R.H.S</p><p></p><p>\frac{m^{2}-1 }{n^{2}-1}$

Replace the value of m & n by the Equation 1 & Equation 2 Respectively.

= \frac{(\frac{sinx}{siny}) ^{2}- 1}{(\frac{tanx}{tany})^{2}-1 }

(

tany

tanx

)

2

−1

(

siny

sinx

)

2

−1

= \frac{\frac{sin^{2}x-sin^{2}y }{sin^{2}y } }{\frac{tan^{2}x -tan^{2}y }{tan^{2} y} }

tan

2

y

tan

2

x−tan

2

y

sin

2

y

sin

2

x−sin

2

y

as we know tan^{2} y = \frac{sin^{2}y }{cos^{2} y}tan

2

y=

cos

2

y

sin

2

y

therefore,

= \frac{\frac{sin^{2}x-sin^{2}y }{sin^{2}y } }{\frac{tan^{2}x - tan^{2}y }{\frac{sin^{2}y}{cos^{2} y}} }

cos

2

y

sin

2

y

tan

2

x− tan

2

y

sin

2

y

sin

2

x−sin

2

y

=\frac{sin^{2}x-sin^{2}y }{sin^{2}y }

sin

2

y

sin

2

x−sin

2

y

\frac{sin^{2}y }{cos^{2}y(tan^{2}x-tan^{2}y)}

cos

2

y(tan

2

x−tan

2

y)

sin

2

y

=\frac{sin^{2}x-sin^{2}y }{cos^{2}y(tan^{2}x-tan^{2}y)}

cos

2

y(tan

2

x−tan

2

y)

sin

2

x−sin

2

y

we know that,

= sin^{2} xsin

2

x =1-cos^{2} x1−cos

2

x

therefore replace sin^{2} xsin

2

x =1-cos^{2} x1−cos

2

x

=\frac{1-cos^{2}x-1+cos^{2}y }{cos^{2}y(tan^{2}x-tan^{2}y)}

cos

2

y(tan

2

x−tan

2

y)

1−cos

2

x−1+cos

2

y

we know that,

=tan^{2} x=sec^{2} x-1tan

2

x=sec

2

x−1

thus,

=\frac{cos^{2}y-cos^{2}x }{cos^{2}y(sec^{2}x-1-sec^{2}y+1)}

cos

2

y(sec

2

x−1−sec

2

y+1)

cos

2

y−cos

2

x

=\frac{cos^{2}y-cos^{2}x }{cos^{2}y(\frac{1}{cos^{2}x } -\frac{1}{cos^{2}y})}

cos

2

y(

cos

2

x

1

−

cos

2

y

1

)

cos

2

y−cos

2

x

=\frac{cos^{2}y-cos^{2}x }{cos^{2}y(\frac{cos^{2}y -cos^{2}x }{cos^{2}x .cos^{2}y})}

cos

2

y(

cos

2

x.cos

2

y

cos

2

y−cos

2

x

)

cos

2

y−cos

2

x

= cos^{2}xcos

2

x =L.H.S