Question

If $\tan \alpha + \cot \alpha = p$ then Prove that ${\tan ^3}\alpha + {\cot ^3}\alpha = p\left( {{p^2} - 3} \right)$.

Answer 1

Hint: Here, we will cube the given equation and apply a suitable algebraic identity to expand the given equation. We will then further simplify the equation using reciprocal trigonometric identity to prove the given equation. A trigonometric equation is defined as an equation involving trigonometric ratios. Trigonometric identity is an equation that is always true.

Formula Used:
 We will use the following formulas:
1.The algebraic Identity for the cube of the sum of the variables is given by ${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$ .
2.Trigonometric ratios: $\tan \alpha = \dfrac{1}{{\cot \alpha }}$ and $\cot \alpha = \dfrac{1}{{\tan \alpha }}$

Complete step-by-step answer:
We are given that
$ \Rightarrow \tan \alpha + \cot \alpha = p$………………………………………………………………………………………………………….$\left( 1 \right)$
Now, cubing the equation $\tan \alpha + \cot \alpha = p$, we get
$ \Rightarrow {\left( {\tan \alpha + \cot \alpha } \right)^3} = {p^3}$
The algebraic Identity for the cube of the sum of the variables is given by ${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$ .
$ \Rightarrow {\tan ^3}\alpha + 3{\tan ^2}\alpha \cot \alpha + 3{\cot ^2}\alpha \tan \alpha + {\cot ^3}\alpha = {p^3}$
We know that $\tan \alpha = \dfrac{1}{{\cot \alpha }}$ and $\cot \alpha = \dfrac{1}{{\tan \alpha }}$
Substituting these trigonometric ratios in the above equation, we get
$ \Rightarrow {\tan ^3}\alpha + 3{\tan ^2}\alpha \cdot \dfrac{1}{{\tan \alpha }} + 3{\cot ^2}\alpha \cdot \dfrac{1}{{\cot \alpha }} + {\cot ^3}\alpha = {p^3}$
By cancelling the similar terms in multiplying the trigonometric ratios, we get
$ \Rightarrow {\tan ^3}\alpha + 3\tan \alpha + 3\cot \alpha + {\cot ^3}\alpha = {p^3}$

By taking out the common factor, we get
$ \Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha + 3\left( {\tan \alpha + \cot \alpha } \right) = {p^3}$
By substituting equation $\left( 1 \right)$ in above equation, we get
$ \Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha + 3\left( p \right) = {p^3}$
By rewriting the equation, we get
$ \Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha = {p^3} - 3p$
By taking out the common factor, we get
$ \Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha = p\left( {{p^2} - 3} \right)$
Therefore, we have proved that ${\tan ^3}\alpha + {\cot ^3}\alpha = p\left( {{p^2} - 3} \right)$and thus verified.

Note: We should know that we have many trigonometric identities which are related to all the other trigonometric equations. We should remember that the trigonometric ratio and the co-trigonometric ratio is always the reciprocal to each other. Trigonometric ratios are used to find the relationships between the sides of a right angle triangle. Whenever squaring or cubing, it has to be done on both sides, to equalize the equation as before.