Q)  Prove that: (1 + cot ² θ) (1 – cos θ) (1 + cos θ) = 1

Ans: 

Method 1:

Step 1: Let’s start with LHS and put values of cot θ:

LHS = (1 + cot ² θ) (1 – cos θ) (1 + cos θ)

= (1 + (  [∵ (a – b) (a + b) = (a² – b² )]

=

Step 2: We know that sin ² θ + cos ² θ = 1,

1 – cos ² θ = sin ² θ

By substituting these values in LHS, we get:

LHS =

=

= 1

Hence Proved !

Method 2:

Step 1: Let’s start with LHS and applying trigonometric identities:

LHS = (1 + cot ² θ) (1 – cos θ) (1 + cos θ)

We know that 1 + cot ² θ = cosec ² θ

∴ LHS = (1 + cot ²  θ) (1 – cos θ) (1 + cos θ)

= cosec ² θ (1 – cos ² θ)

Step 2: We know that sin ² θ + cos ² θ = 1,

1 – cos ² θ = sin ² θ

By substituting these values in LHS, we get:

LHS = cosec ² θ (1 – cos ² θ)

= cosec ² θ (sin ² θ)

=  1

Hence Proved !

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