Tan2A.tanA+1 = tan2A.cotA - 1 = sec2A
Answers
Answered by
17
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
={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
Answered by
3
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
Similar questions