Question

The value of following trigonometric expression \[[{\text{co}}{{\text{s}}^4}{\text{A}} - {\sin ^4}{\text{A}}\]] is equal to:
A. 2${\cos ^2}{\text{A}}$+1
B. 2${\cos ^2}{\text{A}}$−1
C. 2${\sin ^2}{\text{A}}$−1
D. 2${\sin ^2}{\text{A}}$+1

Answer 1

Hint- Proceed the solution of this question, fist identify that there is even power on trigonometric terms so that we can factorise this using available algebraic identity using the difference of square which states as (${{\text{a}}^2} - {{\text{b}}^2}$) = (a+b)(a-b).

Complete step by step answer:
So, by given equation can be written as
⇒\[{\left( {{\text{co}}{{\text{s}}^2}{\text{A}}} \right)^2} - {\left( {{{\sin }^2}{\text{A}}} \right)^2}\]
So on comparing this with the identity difference of square, (${{\text{a}}^2} - {{\text{b}}^2}$) = (a+b)(a-b)
⇒\[{\text{a = co}}{{\text{s}}^2}{\text{A & b = si}}{{\text{n}}^2}{\text{A}}\]
So applying identity we get
⇒ \[{\text{co}}{{\text{s}}^4}{\text{A}} - {\sin ^4}{\text{A}}\] = \[\left( {{\text{co}}{{\text{s}}^2}{\text{A}} - {{\sin }^2}{\text{A}}} \right)\left( {{\text{co}}{{\text{s}}^2}{\text{A + }}{{\sin }^2}{\text{A}}} \right)\]…….(1)
Since we know that \[{\text{co}}{{\text{s}}^2}{\text{A + }}{\sin ^2}{\text{A = 1}}\]
So on putting \[{\text{co}}{{\text{s}}^2}{\text{A + }}{\sin ^2}{\text{A = 1}}\] in equation (1)
⇒\[{\text{co}}{{\text{s}}^4}{\text{A}} - {\sin ^4}{\text{A}}\]= \[{\text{co}}{{\text{s}}^2}{\text{A}} - {\sin ^2}{\text{A}}\]
Now we know that \[{\text{co}}{{\text{s}}^2}{\text{A + }}{\sin ^2}{\text{A = 1}}\], which can be written as \[{\sin ^2}{\text{A = 1 - co}}{{\text{s}}^2}{\text{A}}\]

⇒\[{\text{co}}{{\text{s}}^4}{\text{A}} - {\sin ^4}{\text{A}}\]= \[{\text{co}}{{\text{s}}^2}{\text{A}} - (1 - {\text{co}}{{\text{s}}^2}{\text{A}})\]
On further solving
⇒\[{\text{co}}{{\text{s}}^2}{\text{A}} - (1 - {\text{co}}{{\text{s}}^2}{\text{A}})\] = \[{\text{co}}{{\text{s}}^2}{\text{A}} - 1 + {\text{co}}{{\text{s}}^2}{\text{A = 2co}}{{\text{s}}^2}{\text{A - 1}}\]
⇒\[{\text{co}}{{\text{s}}^4}{\text{A}} - {\sin ^4}{\text{A}}\] \[{\text{ = 2co}}{{\text{s}}^2}{\text{A - 1}}\]

Hence, the option B , 2${\cos ^2}{\text{A}}$−1 is the correct answer.

Note: There is also an alternative approach to solve this question-
Since we know that \[{\text{co}}{{\text{s}}^2}{\text{A + }}{\sin ^2}{\text{A = 1}}\]
which can be written as \[{\sin ^2}{\text{A = 1 - co}}{{\text{s}}^2}{\text{A}}\]
 So replacing the \[{\sin ^2}{\text{A }}\] term by \[{\text{1 - co}}{{\text{s}}^2}{\text{A}}\] in the above question
 \[ \Rightarrow {\text{co}}{{\text{s}}^4}{\text{A}} - {\left( {1 - {{\cos }^2}{\text{A}}} \right)^2}\]
\[ \Rightarrow {\text{co}}{{\text{s}}^4}{\text{A}} - \left( {{\text{co}}{{\text{s}}^4}{\text{A}} + 1 - 2{{\cos }^2}{\text{A}}} \right)\]
\[ \Rightarrow {\text{co}}{{\text{s}}^4}{\text{A}} - {\text{co}}{{\text{s}}^4}{\text{A - }}1 + 2{\cos ^2}{\text{A}}\]
On further solving
\[{\text{ = 2co}}{{\text{s}}^2}{\text{A - 1}}\]