The cosine of angle $x$ minus cosine of angle three times $x$ divided by cosine of angle $x$ plus sine of angle $x$ plus sine of angle three times $x$ divided by sine of angle $x$ is a trigonometric expression in the given trigonometry problem.

$\dfrac{\cos{x}-\cos{3x}}{\cos{x}}$ $+$ $\dfrac{\sin{x}+\sin{3x}}{\sin{x}}$

The trigonometric functions are involved with triple angle in the numerator of both terms of the trigonometric expression. So, the trigonometric expression can be evaluated by the triple angle trigonometric identities.

The second terms in the numerator of both terms in the given trigonometric expression are two trigonometric functions with triple angle. So, they can be expanded as per the triple angle identities.

$(1).\,\,$ $\sin{3x}$ $\,=\,$ $3\sin{x}$ $-$ $4\sin^3{x}$

$(2).\,\,$ $\cos{3x}$ $\,=\,$ $4\cos^3{x}$ $-$ $3\cos{x}$

According to the sine triple angle identity, the sine of angle three times $x$ can be expanded and the cosine of angle three times $x$ can also be expanded as per the cosine triple angle identity.

$=\,\,$ $\dfrac{\cos{x}-(4\cos^3{x}-3\cos{x})}{\cos{x}}$ $+$ $\dfrac{\sin{x}+(3\sin{x}-4\sin^3{x})}{\sin{x}}$

The expression in both numerator and denominator in both terms of the expression is purely expressed in terms of sine of angle $x$ and cosine of angle $x$. Therefore, the expressions in each term can be simplified for finding the value of the trigonometric expression.

$=\,\,$ $\dfrac{\cos{x}-4\cos^3{x}+3\cos{x}}{\cos{x}}$ $+$ $\dfrac{\sin{x}+3\sin{x}-4\sin^3{x}}{\sin{x}}$

$=\,\,$ $\dfrac{\cos{x}+3\cos{x}-4\cos^3{x}}{\cos{x}}$ $+$ $\dfrac{\sin{x}+3\sin{x}-4\sin^3{x}}{\sin{x}}$

$=\,\,$ $\dfrac{4\cos{x}-4\cos^3{x}}{\cos{x}}$ $+$ $\dfrac{4\sin{x}-4\sin^3{x}}{\sin{x}}$

The cosine of angle $x$ is a common factor in both terms of the numerator of first term and the sine of angle $x$ is a common factor in both terms of the numerator of second term in the trigonometric expression. Now, they can be taken out common from the terms.

$=\,\,$ $\dfrac{\cos{x} \times (4-4\cos^2{x})}{\cos{x}}$ $+$ $\dfrac{\sin{x} \times (4-4\sin^2{x})}{\sin{x}}$

$=\,\,$ $\dfrac{\cancel{\cos{x}} \times (4-4\cos^2{x})}{\cancel{\cos{x}}}$ $+$ $\dfrac{\cancel{\sin{x}} \times (4-4\sin^2{x})}{\cancel{\sin{x}}}$

The terms in the trigonometric expression are released from the rational form. Now, let us focus on simplifying the trigonometric expression further.

$=\,\,$ $(4-4\cos^2{x})$ $+$ $(4-4\sin^2{x})$

$=\,\,$ $4-4\cos^2{x}$ $+$ $4-4\sin^2{x}$

$=\,\,$ $4$ $+$ $4$ $-$ $4\cos^2{x}$ $-$ $4\sin^2{x}$

$=\,\,$ $8$ $-$ $4\cos^2{x}$ $-$ $4\sin^2{x}$

The trigonometric expression is simplified and it is time to find the value of the trigonometric expression. Negative four is a common factor in both second and third terms. So, it can be taken out common from them.

$=\,\,$ $8$ $-$ $4 \times (\cos^2{x}+\sin^2{x})$

According to the Pythagorean identity of sine and cosine, the sum of squares of sine and cosine functions is equal to one.

$=\,\,$ $8$ $-$ $4 \times (1)$

$=\,\,$ $8$ $-$ $4 \times 1$

$=\,\,$ $8$ $-$ $4$

$=\,\,$ $4$

$\dfrac{\cos{x}-\cos{3x}}{\cos{x}}$ $+$ $\dfrac{\sin{x}+\sin{3x}}{\sin{x}}$

Learn how to evaluate the $\cos{x}$ minus $\cos{3x}$ divided by $\cos{x}$ plus $\sin{x}$ plus $\sin{3x}$ divided by $\sin{x}$ by the sum and difference to product transformation identities.

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