Question

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

Answer 1

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

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

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