Question

tan a cote
+
1-cote 1-tane
1t sece
29. Prove that​

Answer 1

Answer:

tanA+1−tanAcotA=1+secAcscA

Taking L.H.S.-

1−cotAtanA+1−tanAcotA

=1−(tanA1)tanA+1−tanA(tanA1)

=tanA−1tan2A+tanA(1−tanA)1

=tanA(1−tanA)1−tan3A

=tanA(1−tanA)(1−tanA)(1+tanA+tan2A)(∵a3−b3=(a−b)(a2+ab+b2))

=