Verify the trig equation (:

cos(2x)sec^2(x)csc^2(x)
=
cot^2(x)-tan^2(x)

When verifying trig identities, you CANNOT cross onto both sides. Only work on one side at a time & it should always come out equal.

I've tried this multiple times and I can't seem to get it right. Although, I do know that it is possible. Thanks for your time. Please show your steps.

cos(2x) sec²x csc²x

= cos(2x) * 1/cos²x * 1/sin²x

= (cos²x − sin²x) / (cos²x sin²x)

= (cos²x − sin²x) * 1 / (cos²x sin²x) ----------> replace 1 with cos²x + sin²x

= (cos²x − sin²x) * (cos²x + sin²x) / (cos²x sin²x)

= (cos⁴x − sin⁴x) / (cos²x sin²x)

= cos⁴x/(cos²x sin²x) − sin⁴x/(cos²x sin²x)

= cos²x/sin²x − sin²x/cos²x

= cot²x − tan²x

cos(2x) sec²(x)csc²(x)

equals

(cos²x-sin²x )sec²(x)csc²(x)


equals

(cos²x- sin²x )
▬▬▬▬▬
cos²x sin²x


equals

cos²x ...……….. ..sin²x
▬▬▬▬▬..-..▬▬▬▬▬
cos²x sin²x.......cos²x sin²x

equals

.1...……….. 1
▬▬..-..▬▬▬▬▬
sin²x.......cos²x

equals

csc² x - sec²x

Now it turns out that

1 + tan² x = sec²x

and

1 + cot²x = csc² x

so

csc² x - sec²x

equals

1+ cot²x - (1 + tan²x)

equals

1+ cot²x - 1 - tan²x

equals

cot²x - tan²x

yes. Rick Astley is a mom.

mom