Prove that tan2A = (sec2A+1)√sec²A-1

Prove thattan2A = (sec2A+1)√sec²A-1
1. 3 answers
2. asked by Kewal
3. 1,152 views
Prove that. (1+sec2A) (1+sec4A) (1+sec8A)=Tan8A×CotA
1. 1 answer
2. asked by Bimal
3. 1,017 views
If Tan (A+B)=1/3 and Tan(A-B)=2/5,then prove thatA.Tan2A=11/13 B.Tan2B=-1/17
1. 1 answer
2. asked by Gunda
3. 560 views
If Tan (A+B)=1/3 and Tan(A-B)=2/5,then prove thatA.Tan2A=11/13 B.Tan2B=-1/17
1. 7 answers
2. asked by Rabinprakash Dahal
3. 3,394 views
If Tan (A+B)=1/3 and Tan(A-B)=2/5,then prove thatA.Tan2A=11/13 B.Tan2B=-1/17
1. 2 answers
2. asked by Rabinprakash Dahal
3. 552 views
Prove that(tan4A + tan2A)(1-tan²3Atan²A) = 2tan3Asec²A
1. 2 answers
2. asked by Kewal
3. 810 views
Prove thatsec8A-1/sec4A-1 = tan8A/tan2A
1. 0 answers
2. asked by Kewal
3. 673 views
what is tan2a if sina=1/2
1. 2 answers
2. asked by skorge
3. 2,342 views
Find tan2A when tanA = 1/3 and π<A<3π/2.A) 4/5 B) 5/4 C) 3/4 D) 3/5
1. 1 answer
2. asked by TeeKay
3. 577 views
sec8A-1/sec4A-1=tan8A/tan2A
1. 2 answers
2. asked by Anonymous
3. 1,002 views