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Find $k$ if $1-\dfrac{\sin^2{\alpha}}{1+\cot{\alpha}}$ $-$ $\dfrac{\cos^2{\alpha}}{1+\tan{\alpha}}$ $=$ $k\sin{\alpha}\cos{\alpha}$

Alpha ($\alpha$) is an angle of right triangle and a trigonometric equation is developed in mathematical form by including a constant $k$. The value of $k$ can be calculated by solving this trigonometric equation.

Convert trigonometric functions in quotient form

The numerators of second and third trigonometric terms contain sine and cosine function but the denominators of same terms contain cotangent and tangent functions. So, try to convert both cot and tan functions in terms of sin and cos functions to simplify the trigonometric equation.

It can be done by applying quotient rules of trigonometry.

$\implies$ $1-\dfrac{\sin^2{\alpha}}{1+\dfrac{\cos{\alpha}}{\sin{\alpha}}}$ $-$ $\dfrac{\cos^2{\alpha}}{1+\dfrac{\sin{\alpha}}{\cos{\alpha}}}$ $=$ $k\sin{\alpha}\cos{\alpha}$

Simplify the trigonometric equation

Use basic mathematics to simplify the trigonometric equation easily.

$\implies$ $1-\dfrac{\sin^2{\alpha}}{\dfrac{\sin{\alpha}+\cos{\alpha}}{\sin{\alpha}}}$ $-$ $\dfrac{\cos^2{\alpha}}{\dfrac{\cos{\alpha}+\sin{\alpha}}{\cos{\alpha}}}$ $=$ $k\sin{\alpha}\cos{\alpha}$

$\implies$ $1-\dfrac{\sin^2{\alpha} \times \sin{\alpha}}{\sin{\alpha}+\cos{\alpha}}$ $-$ $\dfrac{\cos^2{\alpha} \times \cos{\alpha}}{\cos{\alpha}+\sin{\alpha}}$ $=$ $k\sin{\alpha}\cos{\alpha}$

$\implies$ $1-\dfrac{\sin^3{\alpha}}{\sin{\alpha}+\cos{\alpha}}$ $-$ $\dfrac{\cos^3{\alpha}}{\sin{\alpha}+\cos{\alpha}}$ $=$ $k\sin{\alpha}\cos{\alpha}$

$\implies$ $1-\Bigg[\dfrac{\sin^3{\alpha}+\cos^3{\alpha}}{\sin{\alpha}+\cos{\alpha}}\Bigg]$ $=$ $k\sin{\alpha}\cos{\alpha}$

Use knowledge of Algebra in Trigonometry

The numerator and denominator contain same trigonometric functions but they contain different exponents. No trigonometric identity can be simplified the second trigonometric term. So, a special mathematical concept should be used to simplify it further and it can be adopted from algebra.

The cube of sum of two terms can be expanded by ${(a+b)}^3$ formula.

${(a+b)}^3$ $=$ $a^3+b^3+3ab(a+b)$

The sum of cubes of two terms can be expressed as two multiplying factors by simplifying it further.

$\implies a^3+b^3$ $=$ ${(a+b)}{(a^2+b^2-ab)}$

Now, take $a = \sin{\alpha}$ and $b = \cos{\alpha}$, then substitute these values in the above algebraic formula.

$\implies \sin^3{\alpha}+\cos^3{\alpha}$ $=$ ${(\sin{\alpha}+\cos{\alpha})}$ ${(\sin^2{\alpha}+\cos^2{\alpha}-\sin{\alpha}\cos{\alpha})}$

$\implies \dfrac{\sin^3{\alpha}+\cos^3{\alpha}}{\sin{\alpha}+\cos{\alpha}}$ $=$ $\sin^2{\alpha}+\cos^2{\alpha}-\sin{\alpha}\cos{\alpha}$

Simplify the trigonometric equation further

$\implies$ $1-\Bigg[\dfrac{\sin^3{\alpha}+\cos^3{\alpha}}{\sin{\alpha}+\cos{\alpha}}\Bigg]$ $=$ $k\sin{\alpha}\cos{\alpha}$

Comeback to simplifying the trigonometric equation. The value of the second trigonometric term is expressed in its equivalent trigonometric expression form. Replace the second term by its equal value.

$\implies 1-$ ${(\sin^2{\alpha}+\cos^2{\alpha}-\sin{\alpha}\cos{\alpha})}$ $=$ $k\sin{\alpha}\cos{\alpha}$

According to Pythagorean identity of sin and cos functions, the value of sum of the squares of sine and cosine at an angle is equal to one.

$\implies 1-$ ${(1-\sin{\alpha}\cos{\alpha})}$ $=$ $k\sin{\alpha}\cos{\alpha}$

$\implies 1-1+\sin{\alpha}\cos{\alpha}$ $=$ $k\sin{\alpha}\cos{\alpha}$

$\implies \require{cancel} \cancel{1}-\cancel{1}+\sin{\alpha}\cos{\alpha}$ $=$ $k\sin{\alpha}\cos{\alpha}$

$\implies \sin{\alpha}\cos{\alpha}$ $=$ $k\sin{\alpha}\cos{\alpha}$

$\implies \dfrac{\sin{\alpha}\cos{\alpha}}{\sin{\alpha}\cos{\alpha}}$ $=$ $k$

$\implies k$ $=$ $\dfrac{\sin{\alpha}\cos{\alpha}}{\sin{\alpha}\cos{\alpha}}$

$\implies k$ $=$ $\require{cancel} \dfrac{\cancel{\sin{\alpha}\cos{\alpha}}}{\cancel{\sin{\alpha}\cos{\alpha}}}$

$\,\,\, \therefore \,\,\,\,\,\, k = 1$

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