As has been pointed out to you before ....
You have posted about 20 of these rather complicated trig identities in the last week or so.
Not once have you shown a single step of your own effort, and not once have you even acknowledged the efforts of tutors to give you detailed and lengthy answers.
We can't even tell if you even look at the answers.
Prove that
tan2A = (sec2A+1)√sec²A-1
3 answers
Tan2A=(sec2A+1)√(sec²A-1)
Tan2a=(sev2a+1)√(sec2a-1)
1 answer