(Q10)
Prove that √
1 + cos θ
1 - cos θ
= cosec θ + cot θ; 0 ≤ θ ≤ 90 ° .
LHS = √
1 + cos θ
1 - cos θ
(multiply numerator and denominator by √(1 +
θ))
= √
1 + cos θ
1 - cos θ
.
1 +
θ
1 +
θ
= √
(1 +
θ)
2
1 -
2
θ
= √
(1 +
θ)
2
2
θ
=
1 +
θ
θ
=
1
θ
+
θ
sin θ
=
θ +
θ = R.H.S.
Clear