If $tanx=-12/5$ , find $((cosx-1)(cosx+1))/((sinx-1)(sinx+1))$ ?

Question

2213 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-29T05:41:19

$((cosx-1)(cosx+1))/((sinx-1)(sinx+1))=144/25$

As $tanx=-12/5$ , we have

$cosx=1/secx=1/sqrt(sec^2x)=1/sqrt(1+tan^2x)$

= $1/sqrt(1+(-12/5)^2)=1/sqrt(1+144/25)=1/sqrt(169/25)$

= $+-1/(13/5)=+-5/13$ and

$sinx=sinx/cosx xxcosx=tanx xxcosx=-12/5xx(+-5/13)=+-12/13$

or $cos^2x=25/169$ and $sin^2x=144/169$

Hence $((cosx-1)(cosx+1))/((sinx-1)(sinx+1))$

= $(cos^2x-1)/(sin^2x-1)=(25/169-1)/(144/169-1)$

= $(-144/169)/(-25/169)=-144/169xx-169/25=144/25$

Answer 2 · 2016-06-29T13:57:02

$144/25$

$((cosx-1)(cosx+1))/((sinx-1)(sinx+1))$

$=-((1-cosx)(1+cosx))/-((1-sinx)(1+sinx))$
$=(1-cos^2x)/(1-sin^2x)$

$=sin^2x/cos^2x=tan^2x=(-12/5)^2=144/25$

If tanx=-12/5tanx=-12/5, find ((cosx-1)(cosx+1))/((sinx-1)(sinx+1))((cosx-1)(cosx+1))/((sinx-1)(sinx+1))?