Question

Prove the following statements.
${\cos ^4}A + {\cot ^2}A = \cos e{c^4}A - \cos e{c^2}A$

Answer 1

Hint: For solving this question, we will first take the common and then by using the trigonometric identities we will be able to prove this statement. Here the formula we will use is $\cos e{c^2}A = 1 + {\cot ^2}A$ . So on substituting this formula we will get to the answer.

Formula used:
Trigonometric identity used is,
$\cos e{c^2}A = 1 + {\cot ^2}A$
Here, $A$ will be the angle

Complete step-by-step answer:
 For solving this type of question we can either go through the LHS or we can go through the RHS. Here in this question we will solve by using the RHS side of the identity.
So by taking the RHS side of the identity, we have
$ \Rightarrow \cos e{c^4}A - \cos e{c^2}A$
So the above identity can also be written as
$ \Rightarrow {\left( {\cos e{c^2}A} \right)^2} - \cos e{c^2}A$
And by using the formula which is given as $\cos e{c^2}A = 1 + {\cot ^2}A$ , we will get the identity as
$ \Rightarrow {\left( {1 + {{\cot }^2}A} \right)^2} - \left( {1 + {{\cot }^2}A} \right)$
And by using the formula of algebra which will be equal to ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , we get
\[ \Rightarrow 1 + 2{\cot ^2}A + {\cot ^4}A - 1 - {\cot ^2}A\]
And on solving it we get the identity as
$ \Rightarrow {\cos ^4}A + {\cot ^2}A$
So here, we can see that the LHS becomes equal to RHS.
Therefore, it is proved that the above statement is correct.
Hence, ${\cos ^4}A + {\cot ^2}A = \cos e{c^4}A - \cos e{c^2}A$ .

Note: For solving this question we can do it by another method also, here the identity used will be the same but the process is little different. Here, we will first use the LHS and by using the identities we will get $\cos e{c^2}A\left( {\cos e{c^2}A - 1} \right)$ . And similarly for the RHS, we will take the ${\cot ^2}A$ common and then by using the identity for the RHS the RHS equation will become the same as the LHS. Hence, from this it is proved that the LHS is equal to the RHS.