# Prove that tanx + cotx = (secx)(cscx).?

### 4 Answers

- Anonymous1 decade agoFavorite Answer
left part

tan x = sinx / cosx

cot x = cosx / sinx

sin²x + cos²x = 1

((sinx)² + (cosx)²) / (cosx*sinx)

= 1 / (cosx* sinx)

right part

sec = 1 / cosx

csec = 1 / sinx

1 / (sinx*cosx)

so left = right

- 1 decade ago
Write everything in terms of sin and cos:

(sinx/cosx)+(cosx/sinx)=(1/cosx)(1/sinx)

Find a common denominator of cosx*sinx on the left side and simplify the right side:

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

Realize that the numerator of the left side is an identity that equals 1. Thus:

1/(sinx*cosx)=1/(sinx*cosx)

- kawaiaeaLv 43 years ago
replace tanx to sinx/cosx then replace cotx to cosx/sinx replace secx to a million/cosx and cscx to a million/sinx sinx/cosx + cosx/sinx=a million/cosx(a million/sinx) while u upload u upload or subtract u could desire to get consumer-friendly denominators for that edge of the equation so situations sinx/cosx by potential of sinx and situations cosx/sinx by potential of cosx sin^2x +cos^2x/cosxsinx=a million/cosx(a million/sinx) sin^2x + cos^2x = a million so replace it for a million a million/cosxsinx=a million/cosx(a million/sinx) so now for the different edge of the equation diverse for the duration of with a million/cosx(a million/sinx) and u get..... a million/cosxsinx=a million/cosxsinx =]

- maegicalLv 41 decade ago
tan x + cot x = sin x/cos x + cos x/sin x =

(sin^2 x/ (cos x *sin x)) + (cos^2 x/(sin x* cos x) =

(sin^2 x + cos^2 x) / (cos x * sin x) =

1/ (cos x * sin x) =

1/ cos x * 1/sin x =

sec x * csc x

Source(s): Geometry class