
Evaluate the expression:
${\text{4}}{\cot ^2}{45^0} - {\sec ^2}{60^0} + {\sin ^2}{60^0} + {\cos ^2}{90^0}$
Answer
552.3k+ views
Hint: Try to evaluate after putting values of trigonometric terms.
Given expression: ${\text{4}}{\cot ^2}{45^0} - {\sec ^2}{60^0} + {\sin ^2}{60^0} + {\cos ^2}{90^0}{\text{ }} \ldots \left( 1 \right)$
Here, we know:
$\cot {45^0} = 1,{\text{ }}\sec {60^0} = 2,{\text{ }}\sin {60^0} = \sqrt {\dfrac{3}{2}} {\text{ and }}\cos {90^0} = 0$
Therefore, putting these values in equation $\left( 1 \right)$, we get
$
{\text{ = 4}}{\left( 1 \right)^2} - {\left( 2 \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + {\left( 0 \right)^2} \\
= 4 - 4 + \left( {\dfrac{3}{4}} \right) + 0 \\
= \dfrac{3}{4} \\
$
Note: Whenever there are angles inside trigonometric functions whose values are known to us, always try to solve by putting the values. Also, try to keep in mind the values of trigonometric functions at some specific angles like ${0^0},{30^0},{45^0},{60^0},{90^0}$as these are used frequently.
Given expression: ${\text{4}}{\cot ^2}{45^0} - {\sec ^2}{60^0} + {\sin ^2}{60^0} + {\cos ^2}{90^0}{\text{ }} \ldots \left( 1 \right)$
Here, we know:
$\cot {45^0} = 1,{\text{ }}\sec {60^0} = 2,{\text{ }}\sin {60^0} = \sqrt {\dfrac{3}{2}} {\text{ and }}\cos {90^0} = 0$
Therefore, putting these values in equation $\left( 1 \right)$, we get
$
{\text{ = 4}}{\left( 1 \right)^2} - {\left( 2 \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + {\left( 0 \right)^2} \\
= 4 - 4 + \left( {\dfrac{3}{4}} \right) + 0 \\
= \dfrac{3}{4} \\
$
Note: Whenever there are angles inside trigonometric functions whose values are known to us, always try to solve by putting the values. Also, try to keep in mind the values of trigonometric functions at some specific angles like ${0^0},{30^0},{45^0},{60^0},{90^0}$as these are used frequently.
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