secx * cosecx - cotx = tanx

We know that:

secx = 1/cosx

csecx = 1/sinx

cot = cosx/sinx

tanx = sinx/cosx

Now substitute:

sec x cosec x - cot x = tan x \

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

1/sinx*cosx - cosx/sinx = sinx/cosx

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

But we know that: sin^2 x + cos^2x = 1

==> 1-cos^2x = sin^2 x

==> sin^2 x /sin*cos = sin/cos

==> sinx /cosx = sinx/cos

==> tanx = tan x

sec x cosec x - cot x = tan x req. to be proved

first of all break all these terms in sin & cos, you will have:

LHS = 1/cos x * 1/sin x - cos x/sin x

= 1/(sin x cos x) - cos x/sin x

= (1-cos2 x)/(cos x sin x)

= sin2 x/cos x sinx

= tan x = RHS

& this is proved dude

To prove secxcosecx -cotx = tanx.

Solution:

LHS : secxcosecx -cotx = 1/cosx*sinx -cosx/sinx

= 1/cosxsinx - cosx*cosx/cosxsinx, converted to acoommon denominator.

= (1-cos^2)/cosxsinx

= sin^2x/cosxsinx, as sin^2x+cos^2x = 1.

= sinx/cosx

= tanx by defintion = RHS

Thus secxcosecx - cotx = tanx.

= (1-