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

Last updated date: 20th Mar 2023

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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 \[\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1\].

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 \[\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1\].

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