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Question: Prove that: \({{\sec }^{6}}A-{{\tan }^{6}}A=1+3{{\tan }^{2}}A+3{{\tan }^{4}}A\)....

Prove that: sec⁡6A−tan⁡6A=1+3tan⁡2A+3tan⁡4A{{\sec }^{6}}A-{{\tan }^{6}}A=1+3{{\tan }^{2}}A+3{{\tan }^{4}}A.

Explanation

Solution

We use the formula of (a)3−(b)3=(a−b)(a2+ab+b2){{\left( a \right)}^{3}}-{{\left( b \right)}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) to factorise the given equation. Then we use the trigonometric identity of (sec⁡2A−tan⁡2A)=1\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)=1. We simplify the values and use sec⁡2A=1+tan⁡2A{{\sec }^{2}}A=1+{{\tan }^{2}}A. We equate both sides of the equation to prove it.

Complete step by step answer:
We have to prove the given equality of sec⁡6A−tan⁡6A=1+3tan⁡2A+3tan⁡4A{{\sec }^{6}}A-{{\tan }^{6}}A=1+3{{\tan }^{2}}A+3{{\tan }^{4}}A.
We take two sides separately and find their simplified answer.
We know (a)3−(b)3=(a−b)(a2+ab+b2){{\left( a \right)}^{3}}-{{\left( b \right)}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)
For the left-hand side equation sec⁡6A−tan⁡6A{{\sec }^{6}}A-{{\tan }^{6}}A, we have
⇒sec⁡6A−tan⁡6A ⇒(sec⁡2A)3−(tan⁡2A)3 ⇒(sec⁡2A−tan⁡2A)(sec⁡4A+sec⁡2Atan⁡2A+tan⁡4A) \begin{aligned} & \Rightarrow {{\sec }^{6}}A-{{\tan }^{6}}A \\\ & \Rightarrow {{\left( {{\sec }^{2}}A \right)}^{3}}-{{\left( {{\tan }^{2}}A \right)}^{3}} \\\ & \Rightarrow \left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)\left( {{\sec }^{4}}A+{{\sec }^{2}}A{{\tan }^{2}}A+{{\tan }^{4}}A \right) \\\ \end{aligned}
We know the trigonometric identity of (sec⁡2A−tan⁡2A)=1\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)=1. Using the property, we get
⇒(sec⁡2A−tan⁡2A)(sec⁡4A+sec⁡2Atan⁡2A+tan⁡4A) ⇒1.(sec⁡4A+sec⁡2Atan⁡2A+tan⁡4A) ⇒(sec⁡2A)2+(sec⁡2A)tan⁡2A+tan⁡4A \begin{aligned} & \Rightarrow \left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)\left( {{\sec }^{4}}A+{{\sec }^{2}}A{{\tan }^{2}}A+{{\tan }^{4}}A \right) \\\ & \Rightarrow 1.\left( {{\sec }^{4}}A+{{\sec }^{2}}A{{\tan }^{2}}A+{{\tan }^{4}}A \right) \\\ & \Rightarrow {{\left( {{\sec }^{2}}A \right)}^{2}}+\left( {{\sec }^{2}}A \right){{\tan }^{2}}A+{{\tan }^{4}}A \\\ \end{aligned}
Now we replace with sec⁡2A=1+tan⁡2A{{\sec }^{2}}A=1+{{\tan }^{2}}A
⇒(sec⁡2A)2+(sec⁡2A)tan⁡2A+tan⁡4A ⇒(1+tan⁡2A)2+(1+tan⁡2A)tan⁡2A+tan⁡4A \begin{aligned} & \Rightarrow {{\left( {{\sec }^{2}}A \right)}^{2}}+\left( {{\sec }^{2}}A \right){{\tan }^{2}}A+{{\tan }^{4}}A \\\ & \Rightarrow {{\left( 1+{{\tan }^{2}}A \right)}^{2}}+\left( 1+{{\tan }^{2}}A \right){{\tan }^{2}}A+{{\tan }^{4}}A \\\ \end{aligned}
Now we simplify the answer by expanding the answer.
⇒(1+tan⁡2A)2+(1+tan⁡2A)tan⁡2A+tan⁡4A ⇒1+2tan⁡2A+tan⁡4A+tan⁡2A+tan⁡4A+tan⁡4A ⇒1+3tan⁡2A+3tan⁡4A \begin{aligned} & \Rightarrow {{\left( 1+{{\tan }^{2}}A \right)}^{2}}+\left( 1+{{\tan }^{2}}A \right){{\tan }^{2}}A+{{\tan }^{4}}A \\\ & \Rightarrow 1+2{{\tan }^{2}}A+{{\tan }^{4}}A+{{\tan }^{2}}A+{{\tan }^{4}}A+{{\tan }^{4}}A \\\ & \Rightarrow 1+3{{\tan }^{2}}A+3{{\tan }^{4}}A \\\ \end{aligned}
Thus, proved R.H.S is equal to L.H.S.

Note: Instead of (a)3−(b)3=(a−b)(a2+ab+b2){{\left( a \right)}^{3}}-{{\left( b \right)}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right), we also could have used the formula of (a)3−(b)3=(a−b)3+3ab(a−b){{\left( a \right)}^{3}}-{{\left( b \right)}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right). Then we had to use the same theorems of (sec⁡2A−tan⁡2A)=1\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)=1 to simplify the answer. We get to prove the same result following the rest of the process in a similar way.