Math Doubts

Evaluate $(x^2y)^\frac{2}{3}-(xy^2)^\frac{2}{3}$, if $x$ $\,=\,$ $\cot{\theta}+\tan{\theta}$ and $y$ $\,=\,$ $\sec{\theta}-\cos{\theta}$

In this trigonometry problem, the values of the variables $x$ and $y$ are expressed in trigonometric functions by considering theta as angle of the right triangle.

$x$ $\,=\,$ $\cot{\theta}+\tan{\theta}$ and $y$ $\,=\,$ $\sec{\theta}-\cos{\theta}$

We have to evaluate the value of an algebraic expression $(x^2y)^\frac{2}{3}-(xy^2)^\frac{2}{3}$ by substituting the values of both variables $x$ and $y$.

Simplify the Expressions of the variables

It is recommendable to simplify the trigonometric expression of each variable before replacing them in the given algebraic expression. Now, let us simplify each equation by some basic trigonometric identities.

$x$ $\,=\,$ $\cot{\theta}+\tan{\theta}$

The cot function can be expressed in tan function form by the reciprocal identity of tan function.

$\implies$ $x$ $\,=\,$ $\dfrac{1}{\tan{\theta}}+\tan{\theta}$

Now, the expression at the right hand side of the equation can be simplified by the addition of the fractions.

$\implies$ $x$ $\,=\,$ $\dfrac{1}{\tan{\theta}}+\dfrac{\tan{\theta}}{1}$

$\implies$ $x$ $\,=\,$ $\dfrac{1 \times 1 + \tan{\theta} \times \tan{\theta}}{\tan{\theta}}$

$\implies$ $x$ $\,=\,$ $\dfrac{1+\tan^2{\theta}}{\tan{\theta}}$

The Pythagorean identity between tan and secant functions can be used to express the trigonometric expression $1+\tan^2{\theta}$ as square of secant function.

$\,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $\dfrac{\sec^2{\theta}}{\tan{\theta}}$

In the same way, the second equation can also be simplified for evaluating the value of variable $y$.

$y$ $\,=\,$ $\sec{\theta}-\cos{\theta}$

Now, use the reciprocal identity of secant function to express it in cosine function.

$\implies$ $y$ $\,=\,$ $\dfrac{1}{\cos{\theta}}-\cos{\theta}$

The right side expression of the equation can be simplified by the subtraction of the fractions.

$\implies$ $y$ $\,=\,$ $\dfrac{1}{\cos{\theta}}-\dfrac{\cos{\theta}}{1}$

$\implies$ $y$ $\,=\,$ $\dfrac{1 \times 1 – \cos{\theta} \times \cos{\theta}}{\cos{\theta}}$

$\implies$ $y$ $\,=\,$ $\dfrac{1-\cos^2{\theta}}{\cos{\theta}}$

According to the Pythagorean identity of sine and cosine functions, the expression $1-\cos^2{\theta}$ can be expressed as square of sine function.

$\,\,\,\therefore\,\,\,\,\,\,$ $y$ $\,=\,$ $\dfrac{\sin^2{\theta}}{\cos{\theta}}$

The given equations $x$ $\,=\,$ $\cot{\theta}+\tan{\theta}$ and $y$ $\,=\,$ $\sec{\theta}-\cos{\theta}$ are simplified as follows.

$\implies$ $x$ $\,=\,$ $\dfrac{\sec^2{\theta}}{\tan{\theta}}$ and $y$ $\,=\,$ $\dfrac{\sin^2{\theta}}{\cos{\theta}}$

Substitute the variables in Algebraic expression

Now, substitute the values of variables $x$ and $y$ in the given algebraic expression $(x^2y)^\frac{2}{3}-(xy^2)^\frac{2}{3}$.

$\implies$ $(x^2y)^{\Large \frac{2}{3}}-(xy^2)^{\Large \frac{2}{3}}$ $\,=\,$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^2{\theta}}{\tan{\theta}}\Bigg)^2\Bigg(\dfrac{\sin^2{\theta}}{\cos{\theta}}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$ $-$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^2{\theta}}{\tan{\theta}}\Bigg)\Bigg(\dfrac{\sin^2{\theta}}{\cos{\theta}}\Bigg)^2\large \Bigg)^{\Large \frac{2}{3}}$

The above equation expresses that the given algebraic expression is completely expressed in the form of trigonometric functions. So, the value of given algebraic expression can be evaluated by simplifying the trigonometric expression.

Evaluate the Algebraic expression by simplification

Now, consider the trigonometric expression for evaluating the given algebraic expression.

$=\,\,\,$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^2{\theta}}{\tan{\theta}}\Bigg)^2\Bigg(\dfrac{\sin^2{\theta}}{\cos{\theta}}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$ $-$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^2{\theta}}{\tan{\theta}}\Bigg)\Bigg(\dfrac{\sin^2{\theta}}{\cos{\theta}}\Bigg)^2\large \Bigg)^{\Large \frac{2}{3}}$

The first factor in the first term and second factor in the second term of the trigonometric expression can be expressed in the below form by the power rule of a quotient.

$=\,\,\,$ $\large \Bigg(\normalsize \Bigg(\dfrac{\Big(\sec^2{\theta}\Big)^2}{(\tan{\theta})^2}\Bigg)\Bigg(\dfrac{\sin^2{\theta}}{\cos{\theta}}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$ $-$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^2{\theta}}{\tan{\theta}}\Bigg)\Bigg(\dfrac{\Big(\sin^2{\theta}\Big)^2}{(\cos{\theta})}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$

The expressions in the numerator of first and second factors in the respective first and second terms can be simplified by the power rule of exponents. In the same way, the expressions in the denominator in the same factors can be written in the below form as per trigonometric mathematics.

$=\,\,\,$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^{2 \times 2}{\theta}}{\tan^2{\theta}}\Bigg)\Bigg(\dfrac{\sin^2{\theta}}{\cos{\theta}}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$ $-$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^2{\theta}}{\tan{\theta}}\Bigg)\Bigg(\dfrac{\sin^{2 \times 2}{\theta}}{\cos^2{\theta}}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^4{\theta}}{\tan^2{\theta}}\Bigg)\Bigg(\dfrac{\sin^2{\theta}}{\cos{\theta}}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$ $-$ $\large \Bigg(\normalsize \Bigg(\dfrac{\sec^2{\theta}}{\tan{\theta}}\Bigg)\Bigg(\dfrac{\sin^4{\theta}}{\cos^2{\theta}}\Bigg)\large \Bigg)^{\Large \frac{2}{3}}$

Now, let us simplify the both terms in the expression by the multiplication of the fractions.

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^4{\theta}}{\tan^2{\theta}} \times \dfrac{\sin^2{\theta}}{\cos{\theta}} \Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\sec^2{\theta}}{\tan{\theta}} \times \dfrac{\sin^4{\theta}}{\cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^4{\theta} \times \sin^2{\theta}}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \sin^4{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

Look at the first term, there is a sine squared function in the numerator and tan squared function in the denominator. They can be cancelled each other by combining the sine squared function with secant squared function. The secant squared function can be taken from the secant raised to the power of four function. In the same way, the sine raised to the power of four function in the second term can be split as product of two sine squared functions as per product rule of exponents. This technique will be useful to us for expressing it in tan squared function, which can be cancelled by the tan function in the denominator.

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^{2+2}{\theta} \times \sin^2{\theta}}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \sin^{2+2}{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \sec^2{\theta} \times \sin^2{\theta}}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \sin^2{\theta} \times \sin^2{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

Now, use reciprocal identity of cos function to express secant squared function in reciprocal form.

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \dfrac{1}{\cos^2{\theta}} \times \sin^2{\theta}}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\dfrac{1}{\cos^2{\theta}} \times \sin^2{\theta} \times \sin^2{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

Multiply sine squared function and cos squared function in reciprocal form for expressing their product in the form of tan function.

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \dfrac{1 \times \sin^2{\theta}}{\cos^2{\theta}}}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\dfrac{1 \times \sin^2{\theta}}{\cos^2{\theta}} \times \sin^2{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \dfrac{\sin^2{\theta}}{\cos^2{\theta}}}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\dfrac{\sin^2{\theta}}{\cos^2{\theta}} \times \sin^2{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \Big(\dfrac{\sin{\theta}}{\cos{\theta}}\Big)^2}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\Big(\dfrac{\sin{\theta}}{\cos{\theta}}\Big)^2 \times \sin^2{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

The quotient of sine by cosine is equal to tangent as per quotient rule of sine and cosine functions.

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times (\tan{\theta})^2}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{(\tan{\theta})^2 \times \sin^4{\theta}}{\tan{\theta} \times \sin^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \tan^2{\theta}}{\tan^2{\theta} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\tan^2{\theta} \times \sin^2{\theta}}{\tan{\theta} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,\require{cancel}$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times \cancel{\tan^2{\theta}}}{\cancel{\tan^2{\theta}} \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\cancel{\tan^2{\theta}} \times \sin^2{\theta}}{\cancel{\tan{\theta}} \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta} \times 1}{1 \times \cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\tan{\theta} \times \sin^2{\theta}}{1 \times \cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

The cancellation part is completed and let’s focus on simplifying each term in the trigonometric expression.

$=\,\,\,$ $\Bigg(\normalsize \dfrac{\sec^2{\theta}}{\cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\normalsize \dfrac{\tan{\theta} \times \sin^2{\theta}}{\cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\dfrac{\sec^2{\theta} \times 1}{\cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\tan{\theta} \times \dfrac{\sin^2{\theta}}{\cos^2{\theta}}\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Bigg(\sec^2{\theta} \times \dfrac{1}{\cos{\theta}}\Bigg)^{\Large \frac{2}{3}}$ $-$ $\Bigg(\tan{\theta} \times \Big(\dfrac{\sin{\theta}}{\cos{\theta}}\Big)^2\Bigg)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Big(\sec^2{\theta} \times \sec{\theta}\Big)^{\Large \frac{2}{3}}$ $-$ $\Big(\tan{\theta} \times (\tan{\theta})^2\Big)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Big(\sec^2{\theta} \times \sec{\theta}\Big)^{\Large \frac{2}{3}}$ $-$ $\Big(\tan{\theta} \times \tan^2{\theta}\Big)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Big(\sec^{2+1}{\theta}\Big)^{\Large \frac{2}{3}}$ $-$ $\Big(\tan^{1+2}{\theta}\Big)^{\Large \frac{2}{3}}$

$=\,\,\,$ $\Big(\sec^3{\theta}\Big)^{\Large \frac{2}{3}}$ $-$ $\Big(\tan^3{\theta}\Big)^{\Large \frac{2}{3}}$

Now, use the power rule of exponents for each term to simplify the trigonometric expression.

$=\,\,\,$ $\sec^{3 \times \Large \frac{2}{3}}{\theta}$ $-$ $\tan^{3 \times \Large \frac{2}{3}}{\theta}$

$=\,\,\,$ $\sec^{\Large \frac{3 \times 2}{3}}{\theta}$ $-$ $\tan^{\Large \frac{3 \times 2}{3}}{\theta}$

$=\,\,\,$ $\sec^{\Large \frac{\cancel{3} \times 2}{\cancel{3}}}{\theta}$ $-$ $\tan^{\Large \frac{\cancel{3} \times 2}{\cancel{3}}}{\theta}$

$=\,\,\,$ $\sec^2{\theta}$ $-$ $\tan^2{\theta}$

According to the Pythagorean identity of secant and tan functions, the value of the trigonometric expression is equal to one.

$=\,\,\,$ $1$

Therefore, it is solved that the value of the algebraic expression $(x^2y)^\frac{2}{3}-(xy^2)^\frac{2}{3}$ is equal to one.

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