How do you verify #(cotx/sec^2x) + (cotx/csc^2x)= cotx#?

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Answer 1 · 2015-11-03T15:54:49

Verify trig expression.
#(cot x/sec^2 x) + (cot x/csc^2 x) = cot x#

First term of left side:
#((cos x/(sin x))/(1/(cos^2 x))) = (cos^3 x)/(sin x) # (1)

Second term of left side:
#((cos x)/sin x)/(1/sin^2 x) = (cos x.sin^2 x) /(sin x)# (2).

Add (1) and (2)
#((cos^3 x)/(sin x))+ (cos x sin^2 x)/(sin x) = #
#= (cos x)/(sin x)(cos^2 x + sin^2 x) = (cos x)/(sin x) = cot x#

Answer 2 · 2015-11-03T16:05:03

Determination is done by following method.

LHS= #cotx/sec^2x+cotx/csc^2x#

#" "color(white)(a)=cotx(1/sec^2x+1/csc^2x)#

#" "color(white)(a)=cotx(cos^2x+sin^2x)#

#" "color(white)(a)=cotx#

#" "color(white)(a)=RHS#