# How do you prove tan(90° + x) = -cot(x)?

We know that tan can be expressed as

tan x = sin x / cos x

tan (90+x) = sin (90+ x) /cos (90+x)——(i)

USing identity

sin(90 + X) = sin(X) cos(90) + cos(X) sin(90)

Also cos(90 + x) = cos(X) cos(90) – sin(X) sin (90)

sin(90) = 1 and cos(90) = 0.

On substituting these in equation (i) we get

$tan (90+x) = \frac{sin(X) cos(90) + cos(X) sin(90)}{cos(X) cos(90) – sin(X) sin (90)}$ $tan (90+x) = \frac{cos x}{ – sin(x)}$

= -cot x