Question 1201964
We could transform the LHS (left hand side) so that it becomes the RHS (right hand side). 
I'll keep the RHS the same throughout each step.


{{{(sin^2(x))/(cos(x)) + sec(x) = (sin(x))/(cot(x)) + 1/cos(x)}}}


{{{(sin(x)*sin(x))/(cos(x)) + sec(x) = (sin(x))/(cot(x)) + 1/cos(x)}}}


{{{sin(x)*((sin(x))/(cos(x))) + 1/cos(x) = (sin(x))/(cot(x)) + 1/cos(x)}}}


{{{sin(x)*tan(x) + 1/cos(x) = (sin(x))/(cot(x)) + 1/cos(x)}}}


{{{sin(x)*(1/cot(x)) + 1/cos(x) = (sin(x))/(cot(x)) + 1/cos(x)}}}


{{{(sin(x))/(cot(x)) + 1/cos(x) = (sin(x))/(cot(x)) + 1/cos(x)}}}


This confirms the original equation is an identity.


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Or we could alter the RHS to transform it into the LHS. 


This time the LHS stays the same for each step.


{{{(sin^2(x))/(cos(x)) + sec(x) = (sin(x))/(cot(x)) + 1/cos(x)}}}


{{{(sin^2(x))/(cos(x)) + sec(x) = sin(x)*(1/cot(x)) + sec(x)}}}


{{{(sin^2(x))/(cos(x)) + sec(x) = sin(x)*tan(x) + sec(x)}}}


{{{(sin^2(x))/(cos(x)) + sec(x) = sin(x)*((sin(x))/(cos(x))) + sec(x)}}}


{{{(sin^2(x))/(cos(x)) + sec(x) = (sin(x)*sin(x))/(cos(x)) + sec(x)}}}


{{{(sin^2(x))/(cos(x)) + sec(x) = (sin^2(x))/(cos(x)) + sec(x)}}}


We get the same thing on both sides, which confirms the equation is an identity.


Here is a list of common identities
<a href = "https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf">https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf</a>
For example, one common identity I used was {{{sec(x) = 1/cos(x)}}}. That example is mentioned under the "Reciprocal Identities" section.