Question

if tanx = ntany and sinx = msiny prove that cos2x = m2-1/n2-1​

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

#LearnMore

https://brainly.in/question/13715501

Answer 2

Question: if tanx = ntany and sinx = msiny, prove that cos^2x = (m^2-1)/(n^2-1)

Hope it helps you