Question

Prove the following statement:
 $2{{\sec }^{2}}\alpha -{{\sec }^{4}}\alpha -2{{\operatorname{cosec}}^{2}}\alpha +{{\operatorname{cosec}}^{4}}\alpha ={{\cot }^{4}}\alpha -{{\tan }^{4}}\alpha $.

Answer 1

Hint: In order to prove the relation given in the question, we should know two basic trigonometric identities, which are, $1+{{\tan }^{2}}\alpha ={{\sec }^{2}}\alpha $ and $1+{{\cot }^{2}}\alpha ={{\operatorname{cosec}}^{2}}\alpha $. We should also know that ${{\left( a+b \right)}^{2}}$ is expanded as $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)$. By using these properties, we can prove the given relation.
Complete step-by-step answer:
In this question, we have been asked to prove that, $2{{\sec }^{2}}\alpha -{{\sec }^{4}}\alpha -2{{\operatorname{cosec}}^{2}}\alpha +{{\operatorname{cosec}}^{4}}\alpha ={{\cot }^{4}}\alpha -{{\tan }^{4}}\alpha $. To prove the same, we will first consider the left hand side or the LHS of the given equation. So, we can write it as, $LHS=2{{\sec }^{2}}\alpha -{{\sec }^{4}}\alpha -2{{\operatorname{cosec}}^{2}}\alpha +{{\operatorname{cosec}}^{4}}\alpha $
Now, we know that ${{\sec }^{2}}\alpha =1+{{\tan }^{2}}\alpha $ and ${{\operatorname{cosec}}^{2}}\alpha =1+{{\cot }^{2}}\alpha $. So, applying that on the LHS, we can write the LHS as follows,
$LHS=2\left( 1+{{\tan }^{2}}\alpha \right)-{{\left( 1+{{\tan }^{2}}\alpha \right)}^{2}}-2\left( 1+{{\cot }^{2}}\alpha \right)+{{\left( 1+{{\cot }^{2}}\alpha \right)}^{2}}$
Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, so we can write ${{\left( 1+{{\tan }^{2}}\alpha \right)}^{2}}=1+{{\tan }^{4}}\alpha +2{{\tan }^{2}}\alpha $ and similarly we can write ${{\left( 1+{{\cot }^{2}}\alpha \right)}^{2}}=1+{{\cot }^{4}}\alpha +2{{\cot }^{2}}\alpha $. So, by substituting these values in the equation on the LHS, we get,
$LHS=2\left( 1+{{\tan }^{2}}\alpha \right)-\left( 1+{{\tan }^{4}}\alpha +2{{\tan }^{2}}\alpha \right)-2\left( 1+{{\cot }^{2}}\alpha \right)+\left( 1+{{\cot }^{4}}\alpha +2{{\cot }^{2}}\alpha \right)$
After simplifying the above equation, we get the LHS as,
$LHS=2+2{{\tan }^{2}}\alpha -1-{{\tan }^{4}}\alpha -2{{\tan }^{2}}\alpha -2-2{{\cot }^{2}}\alpha +1+{{\cot }^{4}}\alpha +2{{\cot }^{2}}\alpha $
Now, we will add the like terms algebraically in the above equation. So, we get the LHS as,
$\begin{align}
  & LHS=\left( 2-2+1-1 \right)+\left( 2{{\tan }^{2}}\alpha -2{{\tan }^{2}}\alpha \right)+\left( {{\cot }^{4}}\alpha \right)+\left( 2{{\cot }^{2}}\alpha -2{{\cot }^{2}}\alpha \right)-{{\tan }^{4}}\alpha \\
 & \Rightarrow LHS=0+0+{{\cot }^{4}}\alpha +0-{{\tan }^{4}}\alpha \\
\end{align}$
We can further simplify and write the above equation as,
$LHS={{\cot }^{4}}\alpha -{{\tan }^{4}}\alpha $ =RHS.
 Therefore, LHS = RHS.
Hence, we have proved the statement given in the question, that is, $2{{\sec }^{2}}\alpha -{{\sec }^{4}}\alpha -2{{\operatorname{cosec}}^{2}}\alpha +{{\operatorname{cosec}}^{4}}\alpha ={{\cot }^{4}}\alpha -{{\tan }^{4}}\alpha $.

Note: There is a high possibility in this question that we may make calculation mistakes. So, we have to be very patient and careful while solving this question. Also we must remember that, $1+{{\tan }^{2}}\alpha ={{\sec }^{2}}\alpha $ and $1+{{\cot }^{2}}\alpha ={{\operatorname{cosec}}^{2}}\alpha $.