Question

Using various trigonometric identities and properties, prove the following results.
${{\cos }^{4}}A-{{\cos }^{2}}A={{\sin }^{4}}A-{{\sin }^{2}}A$.

Answer 1

Hint : We will first transpose ${{\cos }^{4}}A,{{\sin }^{4}}A$ in LHS and ${{\sin }^{2}}A,{{\cos }^{2}}A$ in RHS. We will now solve LHS to show that it is equal to RHS. We use the formula ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ to show that LHS is equal to RHS.

Complete step by step solution :
It is given in the question that we have to prove that ${{\cos }^{4}}A-{{\cos }^{2}}A={{\sin }^{4}}A-{{\sin }^{2}}A$.
We have ${{\cos }^{4}}A-{{\cos }^{2}}A={{\sin }^{4}}A-{{\sin }^{2}}A$
Transposing power 4 terms and power 2 terms to same side, we get,
${{\cos }^{4}}A-{{\sin }^{4}}A={{\cos }^{2}}A-{{\sin }^{2}}A$
Now multiplying both sides with -1, we get,
${{\sin }^{4}}A-{{\cos }^{4}}A={{\sin }^{2}}A-{{\cos }^{2}}A$
Left side of the equation can be modified as follows
${{\left( {{\sin }^{2}}A \right)}^{2}}-{{\left( {{\cos }^{2}}A \right)}^{2}}={{\sin }^{2}}A-{{\cos }^{2}}A$
Now using the general formula, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ on the left side of the equation, we get
$\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)\left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)={{\sin }^{2}}A-{{\cos }^{2}}A$
Since we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$, therefore, we get
$\left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)={{\sin }^{2}}A-{{\cos }^{2}}A$
Therefore, we have proved that $LHS=RHS$.

Note : Students may be bothered by seeing the power of cos and sin as 4. They may directly start searching the formula in trigonometry function with power of 4 to solve this question. But there is no such direct formula including both sin and cos together with power 4 and power 2 in single identity and thus they may skip this question in exam. Therefore, an easy formula is used in the solution to solve this question, with minimal complexity.