# How do you verify the identify #(sintheta+costheta)/sintheta=1+cottheta#?

Recall that

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see explanation.

There are 3 possible approaches that may be taken.

(1) Manipulate the left side to obtain the right side.

(2) Manipulate the right side to obtain the left side.

(3) Manipulate both sides until they are the same.

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To verify the identity ( \frac{\sin(\theta) + \cos(\theta)}{\sin(\theta)} = 1 + \cot(\theta) ), we can start with the left-hand side (LHS) and manipulate it algebraically until it's equivalent to the right-hand side (RHS) of the equation.

Starting with the LHS: [ \frac{\sin(\theta) + \cos(\theta)}{\sin(\theta)} ]

We can rewrite ( \cos(\theta) ) as ( \frac{\sin(\theta)}{\sin(\theta)} ) using the identity ( \cos(\theta) = \frac{\sin(\pi/2 - \theta)}{\sin(\theta)} ): [ = \frac{\sin(\theta) + \frac{\sin(\theta)}{\sin(\theta)}}{\sin(\theta)} ] [ = \frac{\sin(\theta) + \sin(\theta)}{\sin(\theta)} ] [ = \frac{2\sin(\theta)}{\sin(\theta)} ] [ = 2 ]

Now, for the right-hand side (RHS): [ 1 + \cot(\theta) ]

Using the definition of cotangent, ( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} ): [ 1 + \frac{\cos(\theta)}{\sin(\theta)} ]

We can rewrite ( \cos(\theta) ) as ( \frac{\sin(\theta)}{\sin(\theta)} ): [ 1 + \frac{\frac{\sin(\theta)}{\sin(\theta)}}{\sin(\theta)} ] [ = 1 + \frac{\sin(\theta)}{\sin^2(\theta)} ] [ = 1 + \frac{1}{\sin(\theta)} ] [ = \frac{\sin(\theta)}{\sin(\theta)} + \frac{1}{\sin(\theta)} ] [ = \frac{\sin(\theta) + 1}{\sin(\theta)} ]

Therefore, the LHS is equal to the RHS, as both simplify to ( 2 ). Thus, the identity is verified.

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