$$ \frac{\cos heta}{1+\sin heta}=\frac{1-\sin heta}{\cos heta} $$ L.H.S $$ =\frac{\cos heta}{1+\sin heta} $$

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$$ \frac{\cos 	heta}{1+\sin 	heta}=\frac{1-\sin 	heta}{\cos 	heta} $$ L.H.S $$ =\frac{\cos 	heta}{1+\sin 	heta} $$

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We start with the left-hand side (LHS):

LHS = (cos θ)⁄(1 + sin θ)

To show that this equals (1 – sin θ)⁄(cos θ), multiply the numerator and the denominator of the LHS by (1 – sin θ):

LHS = [cos θ · (1 – sin θ)]⁄[(1 + sin θ)(1 – sin θ)]

Recall that (1 + sin θ)(1 – sin θ) is a difference of squares, so:

(1 + sin θ)(1 – sin θ) = 1 – sin² θ

Also, from the Pythagorean identity, we know:

1 – sin² θ = cos² θ

Thus, the expression becomes:

LHS = [cos θ · (1 – sin θ)]⁄(cos² θ)

Cancel one factor of cos θ (provided cos θ ≠ 0) from numerator and denominator:

LHS = (1 – sin θ)⁄(cos θ)

This completes the simplification and shows that:

(cos θ)⁄(1 + sin θ) = (1 – sin θ)⁄(cos θ)

Remember to consider the domain restrictions where cos θ ≠ 0 and 1 + sin θ ≠ 0.
$$ \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta} $$

\( \frac{\cos \theta}{1+\sin \theta}=\frac{1-\sin \theta}{\cos \theta} \) L.H.S \( =\frac{\cos \theta}{1+\sin \theta} \)

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