Question

prove that cos x/sin(90+x)+sin(-x)/sin(180+x) -tan(90+x)/cotx =3

Answer 1

Answer:

$solve\:for\:c,\:\frac{\cos \left(x\right)}{\sin \left(90+x\right)}+\frac{\sin \left(-x\right)}{\sin \left(180+x\right)}-\frac{\tan \left(90+x\right)}{c}otx=3\quad :$

$:\quad c=\frac{xot\tan \left(90+x\right)\sin \left(x+90\right)\sin \left(x+180\right)}{\cos \left(x\right)\sin \left(180+x\right)+\sin \left(-x\right)\sin \left(90+x\right)-3\sin \left(90+x\right)\sin \left(180+x\right)}$

$\cos \left(x\right)\sin \left(180+x\right)+\sin \left(-x\right)\sin \left(90+x\right)-3\sin \left(90+x\right)\sin \left(180+x\right)\ne \:0$

Step-by-step explanation:

$\frac{\cos \left(x\right)}{\sin \left(90+x\right)}+\frac{\sin \left(-x\right)}{\sin \left(180+x\right)}-\frac{\tan \left(90+x\right)}{c}otx=3$

$\frac{\cos \left(x\right)}{\sin \left(90+x\right)}+\frac{\sin \left(-x\right)}{\sin \left(180+x\right)}-\frac{xot\tan \left(90+x\right)}{c}=3$

$\mathrm{Multiply\:by\:LCM=}c\sin \left(x+90\right)\sin \left(x+180\right)$

$\frac{\cos \left(x\right)}{\sin \left(90+x\right)}c\sin \left(x+90\right)\sin \left(x+180\right)+\frac{\sin \left(-x\right)}{\sin \left(180+x\right)}c\sin \left(x+90\right)\sin \left(x+180\right)-$

$\frac{xot\tan \left(90+x\right)}{c}c\sin$  $(x + 90)sin(x+180) = 3csin(x+90)sin(x+180)$

$c\cos \left(x\right)\sin \left(x+180\right)+c\sin \left(-x\right)\sin \left(x+90\right)-xot\tan \left(90+x\right)\sin \left(x+90\right)\sin \left(x+180\right)=3c\sin \left(x+90\right)sin(x+180)$

$\mathrm{Add\:}xot\tan \left(90+x\right)\sin \left(x+90\right)\sin \left(x+180\right)\mathrm{\:to\:both\:sides}$

$c\cos \left(x\right)\sin \left(x+180\right)+c\sin \left(-x\right)\sin \left(x+90\right)-xot\tan \left(90+x\right)\sin \left(x+90\right)\sin \left(x+180\right)$

$xot\tan \left(90+x\right)\sin \left(x+90\right)\sin \left(x+180\right)$ $= 3csin (x+90)sin(x+180)+xottan(90+x)sin(x+90)sin(x+180)$

$c\cos \left(x\right)\sin \left(x+180\right)+c\sin \left(-x\right)\sin \left(x+90\right)=3c\sin \left(x+90\right)\sin \left(x+180\right)+xot\tan \left(90+x\right)\sin \left(x+90\right)sin(x+180)$

$\mathrm{Subtract\:}3c\sin \left(x+90\right)\sin \left(x+180\right)\mathrm{\:from\:both\:sides}$

$c\cos \left(x\right)\sin \left(x+180\right)+c\sin \left(-x\right)\sin \left(x+90\right)-3c\sin \left(x+90\right)\sin \left(x+180\right)=$

$3c\sin \left(x+90\right)\sin \left(x+180\right)$ $+xot\tan \left(90+x\right)\sin \left(x+90\right)\sin \left(x+180\right)-3c\sin \left(x+90\right)\sin \left(x+180\right)$

$c\cos \left(x\right)\sin \left(x+180\right)+c\sin \left(-x\right)\sin \left(x+90\right)-3c\sin \left(x+90\right)\sin \left(x+180\right)=$

$xot\tan \left(90+x\right)\sin \left(x+90\right)$ $sin(x+180)$

$c\left(\cos \left(x\right)\sin \left(180+x\right)+\sin \left(-x\right)\sin \left(90+x\right)-3\sin \left(90+x\right)\sin \left(180+x\right)\right)=$

$xottan(90+x)sin(x+90)sin(x+180)$

$\mathrm{Divide\:both\:sides\:by\:}\cos \left(x\right)\sin \left(180+x\right)+\sin \left(-x\right)\sin \left(90+x\right)-3\sin \left(90+x\right)\sin \left(180+x\right);$

$cos(x)sin(180+x) +sin(-x) sin(90+x) - 3sin(90+x)sin(180+x)\neq 0$