cotx+tanx)(1/sec x +1/cosec x) = cosec+sec x

We know that cot= cos/sin sec=1/cos and cosec=1/sin

Now substitute:

(cosx/sinx + sin/cos)(cox + sin x)

= (cosx)^2+(sinx)^2/(sinx cosx ) (cos x+sinx)

=(1/sinx cos x)(cos x + sin x)

= (cos x+ sinx/sinx cos x)

= 1/sinx + 1/cos x = cosec x+ sec x

LHS :

cot = cosx/sinx , tanx = sinx/cosx, 1/cosecx=sinx and 1/ secx = cosx in

(cot x + tan x)(1/sec x + 1/cosec x) = (cosx/sinx+sinx/cosx)(cosx+sinx)

=cos^2x/sinx +sinx +cosx+sin^2/cosx

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

= 1/sinx -sinx +sinx + cosx + 1/cosx -cosx

=1/sinx + 1/cosx

= cosecx+ secx = RHS.

We know that cosec x = 1/sin x, so sin x = 1/cosec x and

sec x = 1/cos x, so 1/sec x = cos x

cosec x + secx = 1/sin x + 1/cos x

cosec x+secx = (cos x + sin x)/sin x * cos x

1/sec x + 1/cosec x = cos x + sin x

cotx+tanx = cos x/sin x + sin x/cos x

cotx+tanx = [(cosx)^2 + (sinx)^2]/sin x * cos x

From the fundamental formula of trigonometry

(cosx)^2 + (sinx)^2=1

cotx+tanx = 1/sin x * cos x

The given expression will become:

(cos x + sin x)/sin x * cos x=(1/sin x * cos x)*(cos x + sin x)

We'll get:

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