Question

Tan2A.tanA+1 = tan2A.cotA - 1 = sec2A

Answer 1

tan2A.tanA+1
={2tanA/(1-tan²A)}tanA+1
=2tan²A/(1-tan²A)+1
=(2tan²A+1-tan²A)/(1-tan²A)
=(1+tan²A)/(1-tan²A)
=1+(1+tan²A)/(1-tan²A)-1
=(1-tan²A+1+tan²A)/(1-tan²A)-1
=2/(1-tan²A)-1
=(2×tanA×cotA)/(1-tan²A)-1
={2tanA/(1-tan²A)}cotA-1
=tan2A.cotA-1 -------------------------------(1)
tan2A.cotA-1
={2tanA/(1-tan²A)}(1/tanA)-1
=2/(1-tan²A)-1
=(2-1+tan²A)/(1-tan²A)
=(1+tan²A)/(1-tan²A)
={1+(sin²A/cos²A)}/{1-(sin²A/cos²A)}
=(cos²A+sin²A)/(cos²A-sin²A)
=1/(cos²A-sin²A)
=1/cos2A
=sec2A ---------------------------------------(2)
From (1) and (2) we get,
tan2A.tanA+1=tan2A.cotA-1=sec2A 

Answer 2

Answer:

Here, tan2A tanA+1

=2tanA/1-tan^A*tanA+1

=2tan^A/1-tan^A+1

= 2tan^A+1-tan^A/1-tan^A

=tan^A+1/1-tan^A

=sin^A/cos^A+1//1-sin^A/cos^A

=sin^A+cos^A/cos^A-sin^A

=1/cos^A-sin^A

=1/cos2A

=sec2A

And, tan2A. cotA-1

=sin2A/cos2A.cosA/sinA-1

=2sinAcosA/cos^A-sin^A.cosA/sinA-1

=2cosA/cos^A-sin^A.cosA-1

=2cos^A/cos^A-sin^A-1

=2cos^A-cos^A+sin^A/cos^A-sin^A

=cos^A+sin^A/cos2A

=1/cos2A

=sec2A