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

Prove that sec⁡6θ=1+tan⁡6θ+3sec⁡2θtan⁡2θ{\sec ^6}\theta = 1 + {\tan ^{^6}}\theta + 3{\sec ^2}\theta {\tan ^2}\theta .

Explanation

Solution

Hint- Use the algebraic identity (a3−b3=(a−b)(a2+b2+ab))\left( {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right) and (sec⁡2θ−tan⁡2θ)=1\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1.

According to question Let first rewrite the equation given as below-
sec⁡6θ−tan⁡6θ−3sec⁡2θtan⁡2θ=1{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1
Taking the LHS and rearrange the equation as below
sec⁡6θ−tan⁡6θ−3sec⁡2θtan⁡2θ{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta
⇒(sec⁡2θ)3−(tan⁡2θ)3−3sec⁡2θtan⁡2θ\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 a3−b3{a^3} - {b^3} and by algebraic identity we know that a3−b3=(a−b)(a2+b2+ab){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)
So rewriting the equation (1) we get
⇒(sec⁡2θ−tan⁡2θ)(sec⁡4θ+tan⁡4θ+sec⁡2θtan⁡2θ)−3sec⁡2θtan⁡2θ\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\\}
⇒1(sec⁡4θ+tan⁡4θ+sec⁡2θtan⁡2θ)−3sec⁡2θtan⁡2θ\Rightarrow 1\left( {{{\sec }^{^4}}\theta + {{\tan }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } \right) - 3{\sec ^2}\theta {\tan ^{^2}}\theta
⇒sec⁡4θ+tan⁡4θ+sec⁡2θtan⁡2θ−3sec⁡2θtan⁡2θ\Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta + {\sec ^2}\theta {\tan ^2}\theta - 3{\sec ^2}\theta {\tan ^{^2}}\theta
⇒sec⁡4θ+tan⁡4θ−2sec⁡2θtan⁡2θ\Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta - 2{\sec ^2}\theta {\tan ^{^2}}\theta
Or, above can be written as
⇒(sec⁡2θ−tan⁡2θ)2(tan⁡2θ−sec⁡2θ)2\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\\}

⇒(1)2 or (−1)2 ⇒1  \Rightarrow {\left( 1 \right)^2}{\text{ or }}{\left( { - 1} \right)^2} \\\ \Rightarrow 1 \\\

Therefore LHS sec⁡6θ−tan⁡6θ−3sec⁡2θtan⁡2θ=1{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1 or we can saysec⁡6θ=1+tan⁡6θ+3sec⁡2θtan⁡2θ{\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⁡6θ−tan⁡6θ−3sec⁡2θtan⁡2θ{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta to the form a3−b3{a^3} - {b^3}].Remember the trigonometric identity (sec⁡2θ−tan⁡2θ)=1\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1.