Question

Prove the identity ${{\sec }^{2}}\theta \left( {{\sec }^{2}}\theta -2 \right)+1={{\tan }^{4}}\theta$.

Answer 1

For proving the identity given in the above question, we need to consider the LHS of the identity which is equal to ${{\sec }^{2}}\theta \left( {{\sec }^{2}}\theta -2 \right)+1$. We can simplify the LHS of the given identity by using the trigonometric identity given by $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$. By using this trigonometric identity we can substitute ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$ into the LHS of the given identity, which is equal to ${{\sec }^{2}}\theta \left( {{\sec }^{2}}\theta -2 \right)+1$ so as to obtain $\left( {{\tan }^{2}}\theta +1 \right)\left( {{\tan }^{2}}\theta -1 \right)+1$. On applying the algebraic identity given by $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ onto the simplified expression obtained as $\left( {{\tan }^{2}}\theta +1 \right)\left( {{\tan }^{2}}\theta -1 \right)+1$, we will finally obtain the LHS equal to RHS and hence finally the given identity in the above question will be proved.

Complete step by step solution:
The trigonometric identity, to be proved, as given in the above question is written as
$\Rightarrow {{\sec }^{2}}\theta \left( {{\sec }^{2}}\theta -2 \right)+1={{\tan }^{4}}\theta$
Now, let us try to simplify the LHS of the above identity by considering it as below.
$\Rightarrow LHS={{\sec }^{2}}\theta \left( {{\sec }^{2}}\theta -2 \right)+1......\left( i \right)$
Now, we know the trigonometric identity which is given by
$\begin{aligned} & \Rightarrow 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\\ & \Rightarrow {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta ........\left( ii \right) \\\ \end{aligned}$
Substituting the above identity into the equation (i) we get
$\begin{aligned} & \Rightarrow LHS=\left( 1+{{\tan }^{2}}\theta \right)\left( 1+{{\tan }^{2}}\theta -2 \right)+1 \\\ & \Rightarrow LHS=\left( 1+{{\tan }^{2}}\theta \right)\left( {{\tan }^{2}}\theta -1 \right)+1 \\\ & \Rightarrow LHS=\left( {{\tan }^{2}}\theta +1 \right)\left( {{\tan }^{2}}\theta -1 \right)+1 \\\ \end{aligned}$
Now, we can apply the algebraic identity which is given by $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ on the above expression. For this we can observe that $a={{\tan }^{2}}\theta$ and $b=1$. Therefore, on applying this algebraic identity on the above expression, we get
$\begin{aligned} & \Rightarrow LHS={{\left( {{\tan }^{2}}\theta \right)}^{2}}-{{1}^{2}}+1 \\\ & \Rightarrow LHS={{\tan }^{4}}\theta -1+1 \\\ & \Rightarrow LHS={{\tan }^{4}}\theta ........\left( iii \right) \\\ \end{aligned}$
According to the given identity, we have
$\Rightarrow RHS={{\tan }^{4}}\theta .......\left( iv \right)$
From the above equations (iii) and (iv) we can conclude
LHS=RHS

Note: Instead of simplifying the LHS, we can also simplify the RHS in order to prove the given identity. For this, we can substitute ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ from the trigonometric identity $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$ into the RHS which is ${{\tan }^{4}}\theta$ to obtain ${{\left( {{\sec }^{2}}\theta -1 \right)}^{2}}$, which can be expanded using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ to obtain ${{\sec }^{4}}\theta -2{{\sec }^{2}}\theta +1$. Finally, on taking ${{\sec }^{2}}\theta$ common, we will obtain ${{\sec }^{2}}\theta \left( {{\sec }^{2}}\theta -2 \right)+1$ which is equal to LHS.