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Question: If \(\tan \alpha + \cot \alpha = p\) then Prove that \({\tan ^3}\alpha + {\cot ^3}\alpha = p\left( {...

If tan⁡α+cot⁡α=p\tan \alpha + \cot \alpha = p then Prove that tan⁡3α+cot⁡3α=p(p2−3){\tan ^3}\alpha + {\cot ^3}\alpha = p\left( {{p^2} - 3} \right).

Explanation

Solution

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 (a+b)3=a3+3a2b+3ab2+b3{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} .
2.Trigonometric ratios: tan⁡α=1cot⁡α\tan \alpha = \dfrac{1}{{\cot \alpha }} and cot⁡α=1tan⁡α\cot \alpha = \dfrac{1}{{\tan \alpha }}

Complete step-by-step answer:
We are given that
⇒tan⁡α+cot⁡α=p\Rightarrow \tan \alpha + \cot \alpha = p………………………………………………………………………………………………………….(1)\left( 1 \right)
Now, cubing the equation tan⁡α+cot⁡α=p\tan \alpha + \cot \alpha = p, we get
⇒(tan⁡α+cot⁡α)3=p3\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 (a+b)3=a3+3a2b+3ab2+b3{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} .
⇒tan⁡3α+3tan⁡2αcot⁡α+3cot⁡2αtan⁡α+cot⁡3α=p3\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⁡α=1cot⁡α\tan \alpha = \dfrac{1}{{\cot \alpha }} and cot⁡α=1tan⁡α\cot \alpha = \dfrac{1}{{\tan \alpha }}
Substituting these trigonometric ratios in the above equation, we get
⇒tan⁡3α+3tan⁡2α⋅1tan⁡α+3cot⁡2α⋅1cot⁡α+cot⁡3α=p3\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
⇒tan⁡3α+3tan⁡α+3cot⁡α+cot⁡3α=p3\Rightarrow {\tan ^3}\alpha + 3\tan \alpha + 3\cot \alpha + {\cot ^3}\alpha = {p^3}

By taking out the common factor, we get
⇒tan⁡3α+cot⁡3α+3(tan⁡α+cot⁡α)=p3\Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha + 3\left( {\tan \alpha + \cot \alpha } \right) = {p^3}
By substituting equation (1)\left( 1 \right) in above equation, we get
⇒tan⁡3α+cot⁡3α+3(p)=p3\Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha + 3\left( p \right) = {p^3}
By rewriting the equation, we get
⇒tan⁡3α+cot⁡3α=p3−3p\Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha = {p^3} - 3p
By taking out the common factor, we get
⇒tan⁡3α+cot⁡3α=p(p2−3)\Rightarrow {\tan ^3}\alpha + {\cot ^3}\alpha = p\left( {{p^2} - 3} \right)
Therefore, we have proved that tan⁡3α+cot⁡3α=p(p2−3){\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.