Question

# Prove that ${\sec ^6}\theta = 1 + {\tan ^{^6}}\theta + 3{\sec ^2}\theta {\tan ^2}\theta$.

Hint- Use the algebraic identity $\left( {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right)$ and $\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1$.

According to question Let first rewrite the equation given as below-
${\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1$
Taking the LHS and rearrange the equation as below
${\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta$
$\Rightarrow {\left( {{{\sec }^2}\theta } \right)^3} - {\left( {{{\tan }^2}\theta } \right)^3} - 3{\sec ^2}\theta {\tan ^2}\theta$ ………. (1)
Now if we clearly look the above equation it is of the form ${a^3} - {b^3}$ and by algebraic identity we know that ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
So rewriting the equation (1) we get
$\Rightarrow \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {{{\sec }^{^4}}\theta + {{\tan }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } \right) - 3{\sec ^2}\theta {\tan ^{^2}}\theta$
Now by trigonometric identity we know that $\left\{ {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right\}$
$\Rightarrow 1\left( {{{\sec }^{^4}}\theta + {{\tan }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } \right) - 3{\sec ^2}\theta {\tan ^{^2}}\theta$
$\Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta + {\sec ^2}\theta {\tan ^2}\theta - 3{\sec ^2}\theta {\tan ^{^2}}\theta$
$\Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta - 2{\sec ^2}\theta {\tan ^{^2}}\theta$
Or, above can be written as
$\Rightarrow {\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right)^2}{\left( {{{\tan }^{^2}}\theta - {{\sec }^2}\theta } \right)^2}$
Again, by trigonometric identity we know that $\left\{ {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right\}$
$\ \Rightarrow {\left( 1 \right)^2}{\text{ or }}{\left( { - 1} \right)^2} \\ \Rightarrow 1 \\$
Therefore LHS ${\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1$ or we can say${\sec ^6}\theta = 1 + {\tan ^{^6}}\theta + 3{\sec ^2}\theta {\tan ^2}\theta$
Hence Proved.

Note- Whenever this type of question appears always bring the RHS terms to LHS and then solve LHS. Afterwards rearrange the equation such that it makes an algebraic identity [ as in our question we converted the equation ${\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta$ to the form ${a^3} - {b^3}$].Remember the trigonometric identity $\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1$.