Question

cosec(90+x)+cot(90+x)/cosec(90-x)+tan(180-x)+tan(180+x)+sec(180-x)/tan(360+x)-sec(-x)=2​

Answer 1

Answer:

Step-by-step explanation:

Cosec(90+x)+cot(90+x)/cosec(90-x)+tan(180-x)+tan(180+x)+sec(180-x)/tan(360+x)-sec(-x)=2​

{Cosec(90+x)+cot(90+x)}/{cosec(90-x)+tan(180-x)}  + {tan(180+x)+sec(180-x)}/{tan(360+x)-sec(-x)}

Solving both terms separately

{Cosec(90+x)+cot(90+x)}/{cosec(90-x)+tan(180-x)}

Cosec(90+x) = 1/Sin(90 +x) =  1/Cosx

cot(90+x) = Cos(90 +x)/Sin(90 +x) =  -Sinx/Cosx

cosec(90-x) = 1/Sin(90-x) = 1/Cosx

tan(180-x) = Sin(180-x)/Cos(180-x)  = Sinx/-Cosx  = -Sinx/Cosx

{Cosec(90+x)+cot(90+x)}/{cosec(90-x)+tan(180-x)}

= (1/Cosx - Sinx/Cosx)/(1/Cosx - Sinx/Cosx)

= 1

{tan(180+x)+sec(180-x)}/{tan(360+x)-sec(-x)}

tan(180+x) = Sin(180 +x)/Cos(180 +x)  = -Sinx /-Cosx  = Tanx

sec(180-x) = 1/Cos(180-x) = -1/Cosx

{tan(360+x) = Tanx

sec(-x) = 1/Cos(-x) = 1/Cosx

= (Tanx -  -1/Cosx)/(Tanx - 1/Cosx)

= 1

1 + 1 =2

= RHS

Answer 2

Solution:

As we know that on horizontal axis trigonometric formula does not change,but on vertical axis they changes.Always remember that in 1st Quadrant All positive,2nd Quadrant sin and Cosec are positive,
in 3rd Quadrant tan and cot are positive and in 4th Quadrant cos and sec are positive.
$cosec(90 + x) = sec \: x \\ \\ cot(90 + x) = - tan \: x \\ \\ cosec(90 - x) = sec \: x \\ \\ tan(180 - x) = - tan \: x$
$tan(180 + x) = tan \: x \\ \\ sec(180 - x) = - sec \: x \\ \\ \tan(360 + x) = tan \: x \\ \\ sec( - x) = sec \: x$
put all these values

$\frac{cosec(90 + x) + cot(90 + x)}{cosec(90 - x) + tan(180 - x)} + \frac{tan(180 + x) + sec(180 - x)}{tan(360 + x) + sec(- x)} \\ \\ put \: values \: written \: above \\ \\ = > \frac{sec \: x - tan \: x}{sec \: x - tan \: x} + \frac{tan \: x - sec \: x}{tan \: x - sec \: x} \\ \\ \\ = > 1 + 1 \\ \\ = > 2$
LHS=RHS

hence proved.