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Evaluate $\dfrac{1+\cos{\theta}-\sin^2{\theta}}{\sin{\theta}(1+\cos{\theta})}$

The trigonometric functions sine and cosine in terms of theta form the following trigonometric expression in ratio form. In this trigonometry problem, we learn how to use the basic trigonometric identities to evaluate the given trigonometric expression.


Transform the Square of Sine function

Look at the expressions in the numerator and denominator of the rational trigonometric expression. The expression in the denominator is in simplified form and the expression the numerator is not simplified. Hence, we have to focus on simplifying the trigonometric expression in the numerator firstly.

The first two terms in the expression of the numerator cannot be simplified but the third term can be transformed into another form by the sine squared identity.

$=\,\,\,$ $\dfrac{1+\cos{\theta}-(1-\cos^2{\theta})}{\sin{\theta}(1+\cos{\theta})}$

$=\,\,\,$ $\dfrac{1+\cos{\theta}-1+\cos^2{\theta}}{\sin{\theta}(1+\cos{\theta})}$

$=\,\,\,$ $\dfrac{1-1+\cos{\theta}+\cos^2{\theta}}{\sin{\theta}(1+\cos{\theta})}$

$=\,\,\,\require{cancel}$ $\dfrac{\cancel{1}-\cancel{1}+\cos{\theta}+\cos^2{\theta}}{\sin{\theta}(1+\cos{\theta})}$

$=\,\,\,$ $\dfrac{\cos{\theta}+\cos^2{\theta}}{\sin{\theta}(1+\cos{\theta})}$

Take the Common factor out from the Expression

In the numerator of the rational function, there is a common factor in the both terms of the expression and the common factor is $\cos{\theta}$ and it can be taken out from the trigonometric expression.

$=\,\,\,$ $\dfrac{\cos{\theta} \times 1+\cos^2{\theta}}{\sin{\theta}(1+\cos{\theta})}$

$=\,\,\,$ $\dfrac{\cos{\theta} \times (1+\cos{\theta})}{\sin{\theta}(1+\cos{\theta})}$

$=\,\,\,$ $\dfrac{\cos{\theta}(1+\cos{\theta})}{\sin{\theta}(1+\cos{\theta})}$

Evaluate the Quotient of the Trigonometric function

The trigonometric expression in both numerator and denominator of the expression consist of same factor and they can be cancelled mathematically. Thus, the simplification of the trigonometric expression in the numerator is completed.

$=\,\,\,$ $\dfrac{\cos{\theta}\cancel{(1+\cos{\theta})}}{\sin{\theta}\cancel{(1+\cos{\theta})}}$

$=\,\,\,$ $\dfrac{\cos{\theta}}{\sin{\theta}}$

Now, we can use the cosine by sine quotient identity to evaluate the trigonometric expression.

$=\,\,\,$ $\cot{\theta}$

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