$A$ is an angle of a right angled triangle. $\cos{A}$ and $\sin{A}$ are trigonometric functions and $\cos{3A}$ and $\sin{3A}$ are also trigonometric functions with triple angle. The four functions are formed a trigonometric expression.

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

The trigonometric expression can be simplified to find its value.

01

Cosine triple angle and sine triple angle terms can be expanded by applying the triple angle trigonometric identities.

According to expansion of the sin triple angle formula and cos triple angle formula.

$\sin{3A} \,=\, 3\sin{A}-4\sin^3{A}$

$\cos{3A} \,=\, 4\cos^3{A}-3\cos{A}$

Use the above two formulas and simplify the expression.

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

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

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

02

There is a $\cos{A}$ term in denominator and a $\cos{A}$ term can be taken common from both terms of the numerator in the first term. Similarly, there is a $\sin{A}$ tem in denominator and a $\sin{A}$ term can also be taken common from the both terms in the numerator in the second term.

$=\,\,\,$ $\dfrac{4\cos{A}(1-\cos^2{A})}{\cos{A}}$ $+$ $\dfrac{4\sin{A}(1-\sin^2{A})}{\sin{A}}$

$=\,\,\,$ $\require{cancel} \dfrac{4\cancel{\cos{A}}(1-\cos^2{A})}{\cancel{\cos{A}}}$ $+$ $\require{cancel} \dfrac{4\cancel{\sin{A}}(1-\sin^2{A})}{\cancel{\sin{A}}}$

$=\,\,\,$ $4(1-\cos^2{A})$ $+$ $4(1-\sin^2{A})$

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

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

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

03

As per the Pythagorean identity of sine and cosine functions, the sum of squares of sine and cosine of an angle is one.

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

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

$=\,\,\,$ $4$

It is the required result of this trigonometry problem.

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