Question

Which one of the following options is correct?
A. \[{\sin ^2}{30^ \circ },{\sin ^2}{45^ \circ },{\sin ^2}{60^ \circ }\] are in GP
B. \[{\cos ^2}{30^ \circ },{\cos ^2}{45^ \circ },{\cos ^2}{60^ \circ }\] are in GP
C. \[{\tan ^2}{30^ \circ },{\tan ^2}{45^ \circ },{\tan ^2}{60^ \circ }\] are in GP
D. None of these

Answer 1

Hint: In order to solve this problem, we will apply the formula \[{b^2} = ac\] in every option. And then put the values of trigonometric angles from the trigonometric table.
After doing this, we observe that which option correctly fit in the formula \[{b^2} = ac\].

Formula Used: 1) We will use \[{b^2} = ac\] formula, to check that the given numbers are in G.P. or not.
2) We will also use the values of trigonometric angles from the trigonometric table:
\[\sin {30^ \circ } = \dfrac{1}{2}\] , \[\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]and \[\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\]
\[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\] , \[\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]and \[\cos {60^ \circ } = \dfrac{1}{2}\]
\[\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\] , \[\tan {45^ \circ } = 1\]and \[\tan {60^ \circ } = \sqrt 3 \]

Complete step by step solution: Remember that three numbers \[a\],\[b\], and \[c\]will be in G.P. if \[{b^2} = ac\].
First, we will check \[{\sin ^2}{30^ \circ },{\sin ^2}{45^ \circ },{\sin ^2}{60^ \circ }\] are in G.P. or not.
\[{\left( {{{\sin }^2}{{45}^ \circ }} \right)^2} = {\sin ^2}{30^ \circ } \times {\sin ^2}{60^ \circ }\]
Substitute the values of trigonometric angles
\[ \Rightarrow {\left[ {{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \right]^2} = {\left( {\dfrac{1}{2}} \right)^2} \times {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}\]
\[ \Rightarrow {\left[ {\dfrac{1}{2}} \right]^2} = \dfrac{1}{4} \times \dfrac{3}{4}\]
\[ \Rightarrow \dfrac{1}{4} \ne \dfrac{3}{{16}}\]
We can observe that\[{{\mathop{\rm Sin}\nolimits} ^2}{30^ \circ },{{\mathop{\rm Sin}\nolimits} ^2}{45^ \circ },{{\mathop{\rm Sin}\nolimits} ^2}{60^ \circ }\]are not in GP.
Now we will check \[{\cos ^2}{30^ \circ },{\cos ^2}{45^ \circ },{\cos ^2}{60^ \circ }\]are in G.P. or not.
\[{\left( {{{\cos }^2}{{45}^ \circ }} \right)^2} = {\cos ^2}{30^ \circ } \times {\cos ^2}{60^ \circ }\]
Substitute the values of trigonometric angles
\[ \Rightarrow {\left[ {{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \right]^2} = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} \times {\left( {\dfrac{1}{2}} \right)^2}\]
\[ \Rightarrow {\left[ {\dfrac{1}{2}} \right]^2} = \dfrac{3}{4} \times \dfrac{1}{4}\]
\[ \Rightarrow \dfrac{1}{4} \ne \dfrac{3}{{16}}\]
We can observe that\[{\cos ^2}{30^ \circ },{\cos ^2}{45^ \circ },{\cos ^2}{60^ \circ }\]are not in GP.
Further, we will check \[{\tan ^2}{30^ \circ },{\tan ^2}{45^ \circ },{\tan ^2}{60^ \circ }\]are in G.P. or not.
\[{\left( {{{\tan }^2}{{45}^ \circ }} \right)^2} = {\tan ^2}{30^ \circ } \times {\tan ^2}{60^ \circ }\]
Substitute the values of trigonometric angles
\[ \Rightarrow {\left[ {{1^2}} \right]^2} = {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} \times {\left( {\sqrt 3 } \right)^2}\]
\[ \Rightarrow 1 = \dfrac{1}{3} \times 3\]
\[ \Rightarrow 1 = 1\]
As a result, we can say that\[{\tan ^2}{30^ \circ },{\tan ^2}{45^ \circ },{\tan ^2}{60^ \circ }\]are in GP.

Option ‘C’ is correct

Note: It is important to develop intuition so that we can possibly gain a sense of what might be the most likely answer to a problem.