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Question: How do you simplify \({{\sec }^{4}}s-{{\tan }^{2}}s={{\tan }^{4}}s+{{\sec }^{2}}s\)?...

How do you simplify sec⁡4s−tan⁡2s=tan⁡4s+sec⁡2s{{\sec }^{4}}s-{{\tan }^{2}}s={{\tan }^{4}}s+{{\sec }^{2}}s?

Explanation

Solution

For simplifying the equation given in the above question, we need to separate secant terms on the LHS and the tangent terms on the RHS. For this, we need to subtract sec⁡2s{{\sec }^{2}}s from both the sides of the given equation, and then add tan⁡2s{{\tan }^{2}}s on both the sides. Then, using the trigonometric identity 1+tan⁡2s=sec⁡2s1+{{\tan }^{2}}s={{\sec }^{2}}s, the given equation will become simplified.

Complete step by step answer:
The trigonometric equation given in the above question is written as
⇒sec⁡4s−tan⁡2s=tan⁡4s+sec⁡2s\Rightarrow {{\sec }^{4}}s-{{\tan }^{2}}s={{\tan }^{4}}s+{{\sec }^{2}}s
On subtracting sec⁡2s{{\sec }^{2}}s from both the sides of the above equation, we get
⇒sec⁡4s−tan⁡2s−sec⁡2s=tan⁡4s+sec⁡2s−sec⁡2s ⇒sec⁡4s−tan⁡2s−sec⁡2s=tan⁡4s \begin{aligned} & \Rightarrow {{\sec }^{4}}s-{{\tan }^{2}}s-{{\sec }^{2}}s={{\tan }^{4}}s+{{\sec }^{2}}s-{{\sec }^{2}}s \\\ & \Rightarrow {{\sec }^{4}}s-{{\tan }^{2}}s-{{\sec }^{2}}s={{\tan }^{4}}s \\\ \end{aligned}
Now, on adding tan⁡2s{{\tan }^{2}}s on both the sides of the above equation, we get
⇒sec⁡4s−tan⁡2s−sec⁡2s+tan⁡2s=tan⁡4s+tan⁡2s ⇒sec⁡4s−sec⁡2s=tan⁡4s+tan⁡2s \begin{aligned} & \Rightarrow {{\sec }^{4}}s-{{\tan }^{2}}s-{{\sec }^{2}}s+{{\tan }^{2}}s={{\tan }^{4}}s+{{\tan }^{2}}s \\\ & \Rightarrow {{\sec }^{4}}s-{{\sec }^{2}}s={{\tan }^{4}}s+{{\tan }^{2}}s \\\ \end{aligned}
Now, we can take sec⁡2s{{\sec }^{2}}s common on the LHS of the above equation to get
⇒sec⁡2s(sec⁡2s−1)=tan⁡4s+tan⁡2s\Rightarrow {{\sec }^{2}}s\left( {{\sec }^{2}}s-1 \right)={{\tan }^{4}}s+{{\tan }^{2}}s
Similarly we can take tan⁡2s{{\tan }^{2}}s common on the RHS of the above equation to get
⇒sec⁡2s(sec⁡2s−1)=tan⁡2s(tan⁡2s+1)........(i)\Rightarrow {{\sec }^{2}}s\left( {{\sec }^{2}}s-1 \right)={{\tan }^{2}}s\left( {{\tan }^{2}}s+1 \right)........\left( i \right)
Now, we know the trigonometric identity which is given by
⇒1+tan⁡2s=sec⁡2s........(ii)\Rightarrow 1+{{\tan }^{2}}s={{\sec }^{2}}s........\left( ii \right)
On subtracting 11 from both sides of the above identity, we get

& \Rightarrow 1+{{\tan }^{2}}s-1={{\sec }^{2}}s-1 \\\ & \Rightarrow {{\sec }^{2}}s-1={{\tan }^{2}}s........\left( iii \right) \\\ \end{aligned}$$ On substituting the equations (ii) and (iii) into the equation (i) we get $\begin{aligned} & \Rightarrow {{\sec }^{2}}s\left( {{\tan }^{2}}s \right)={{\tan }^{2}}s\left( {{\sec }^{2}}s \right) \\\ & \Rightarrow {{\sec }^{2}}s{{\tan }^{2}}s={{\tan }^{2}}s{{\sec }^{2}}s \\\ \end{aligned}$ The LHS is clearly equal to the RHS, as can be seen in the above equation. **Note:** For solving these kinds of questions, we need to remember the basic trigonometric identities, which can be easily derived using the most basic identity given by ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. We can also substitute ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ and ${{\sec }^{4}}x={{\left( 1+{{\tan }^{2}}x \right)}^{2}}$ in the given equation so as to obtain it in terms of $\tan x$ only which can be easily simplified by expanding the binomials.