Question

Prove the trigonometric equation ${{\sec }^{6}}A-{{\tan }^{6}}A=1+3{{\tan }^{2}}A{{\sec }^{2}}A$

Answer 1

Use the property of exponent: ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ (Think of powers of 6 as powers of 2 raised to powers of 3). Expand ${{\sec }^{6}}A-{{\tan }^{6}}A$ using the algebraic expansion formula of ${{a}^{3}}-{{b}^{3}}$ and then simplify using the trigonometric identity ${{\sec }^{2}}A-{{\tan }^{2}}A=1$.

Complete step-by-step answer :
We need to prove that ${{\sec }^{6}}A-{{\tan }^{6}}A=1+3{{\tan }^{2}}A{{\sec }^{2}}A$.
First we consider the Left Hand Side (LHS) of the equation to be proved.$LHS={{\sec }^{6}}A-{{\tan }^{6}}A$
Using the property of exponent, LHS can be rewritten as follows,
$LHS={{\sec }^{6}}A-{{\tan }^{6}}A=\,\,{{({{\sec }^{2}}A)}^{3}}-{{({{\tan }^{2}}A)}^{3}}...............(1)$
Now we have LHS in the form$\,{{(a)}^{3}}-{{(b)}^{3}}$
We know the identity,
${{(a-b)}^{3}}={{(a)}^{3}}-{{(b)}^{3}}-3ab(a-b)$
Rearranging this identity we get,
${{(a)}^{3}}-{{(b)}^{3}}={{(a-b)}^{3}}+3ab(a-b)$
Now suppose $a={{\sec }^{2}}A,\,\,b={{\tan }^{2}}A$
${{(a)}^{3}}-{{(b)}^{3}}={{({{\sec }^{2}}A)}^{3}}-{{({{\tan }^{2}}A)}^{3}}$ ${{\sec }^{6}}A-{{\tan }^{6}}A={{({{\sec }^{2}}A-{{\tan }^{2}}A)}^{3}}+3{{\sec }^{2}}A{{\tan }^{2}}A({{\sec }^{2}}A-{{\tan }^{2}}A)$
Applying this in equation (1) we get,
$LHS=\,{{({{\sec }^{2}}A)}^{3}}-{{({{\tan }^{2}}A)}^{3}}$
$={{({{\sec }^{2}}A-{{\tan }^{2}}A)}^{3}}+3{{\sec }^{2}}A{{\tan }^{2}}A({{\sec }^{2}}A-{{\tan }^{2}}A)..........(2)$
We also know the following trigonometric identity,
${{\sec }^{2}}A-{{\tan }^{2}}A=1.............................................\left( 3 \right)$
Using equation (3) in equation (2), we get,
LHS=\,{{\left\\{ 1 \right\\}}^{3}}+3{{\sec }^{2}}A{{\tan }^{2}}A\left\\{ 1 \right\\}
$=\,1+3{{\sec }^{2}}A{{\tan }^{2}}A$
$=RHS$
Hence the required trigonometric equation is proved.

Note : Alternate methods:
Substitute ${{\sec }^{6}}A$ in LHS as ${{\left( {{\sec }^{2}}A \right)}^{3}}$. Now we can use the trigonometric identity ${{\sec }^{2}}A-{{\tan }^{2}}A=1$ to substitute ${{\sec }^{2}}A$ as $1+{{\tan }^{2}}A$. Then expanding ${{\left( 1+{{\tan }^{2}}A \right)}^{3}}$ using the algebraic identity for ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$, we get the desired RHS.
This method can also be done by substituting ${{\tan }^{6}}A$ in LHS. Substitute ${{\tan }^{6}}A$ in LHS as ${{\left( {{\tan }^{2}}A \right)}^{3}}$. Now we can use the trigonometric identity ${{\sec }^{2}}A-{{\tan }^{2}}A=1$ to substitute ${{\tan }^{2}}A$ as ${{\sec }^{2}}A-1$. Then expanding ${{\left( {{\sec }^{2}}A-1 \right)}^{3}}$ using the algebraic identity for ${{\left( a-b \right)}^{3}}$, we get the desired RHS.
One must be careful not to make unnecessary substitutions. Unnecessary substitutions might make the conversion from LHS to RHS more tedious. Make trigonometric substitutions only as per the requirement of the question. In some questions RHS to LHS conversion might be easier, whereas in some others we might need to simplify both LHS and RHS to prove their equivalence. However, in this question LHS to RHS conversion is the easiest.