(Q2)Prove that
sin θ - cos θ+1
sin θ +cos θ-1
 =
1
sec θ - tan θ
 using identity sec2θ=1+tan 2θ

Answer:

L.H.S=
sin θ - cos θ+1
sin θ +cos θ-1
 =
1
sec θ - tan θ

On dividing Nr & Dr by cos θ, we get

=
sin θ
cos θ
 -
cos θ
cos θ
+
1
cos θ
sin θ
cos θ
+
cos θ
cos θ
 -
1
cos θ

=
θ+sec θ - 1
θ - sec θ + 1

[2θ - tan2θ=1]

=
( θ+sec θ) - (2θ - tan2θ)
θ - sec θ + 1

[a2-b2=(a+b)(a-b)]

=
( θ+sec θ) - [( θ+tanθ)( θ-tan θ)]
θ - sec θ + 1

=
( θ+sec θ)(1- θ+tan θ)
θ - sec θ + 1
 = θ + sec θ ---------(1)

R.H.S=
1
sec θ - tan θ

[By rationalising R.F of sec θ - tan θ is sec θ + tan θ]

=
1
sec θ - tan θ
×
sec θ + tan θ
sec θ + tan θ

=
sec θ + tan θ
2 θ - tan2θ

[2θ - tan2θ =1]

= sec θ + tan θ --------(2)

from (1) and (2),we get L.H.S=R.H.S
Hence proved