
If \[7{\cos ^2}{\rm{\theta }} + 3{\sin ^2}{\rm{\theta }} = 4\], then \[\cot {\rm{\theta }} = \]
A.\[7\]
B.\[\dfrac{7}{3}\]
C.\[\sqrt 3 \]
D.\[\dfrac{1}{{\sqrt 3 }}\]
Answer
561.3k+ views
Hint: Here, we have to use the basic identities of the trigonometric functions to find out the value of the given equation i.e. \[\cot {\rm{\theta }}\]. So we have to apply the properties of the trigonometric function for the simplification of the equation and by solving the simplified equation we will get the value of the theta and by putting the value of theta in \[\cot {\rm{\theta }}\] we will get the answer.
Complete step-by-step answer:
Given equation is \[7{\cos ^2}{\rm{\theta }} + 3{\sin ^2}{\rm{\theta }} = 4\]………… (1)
And we have to find out the value of \[\cot {\rm{\theta }}\].
So, firstly we have to use the basic identity of the trigonometric functions to simplify the equation to find the value of the theta.
Here, we know that \[{\rm{si}}{{\rm{n}}^2}{\rm{\theta }} + {\cos ^2}{\rm{\theta = 1}}\], we can also write it as \[{\rm{si}}{{\rm{n}}^2}{\rm{\theta = 1}} - {\cos ^2}{\rm{\theta }}\] . So, we will put the value of the \[{\sin ^2}{\rm{\theta }}\] in the main equation i.e. equation (1), we get
\[ \Rightarrow 7{\cos ^2}{\rm{\theta }} + 3({\rm{1}} - {\cos ^2}{\rm{\theta )}} = 4\]
\[ \Rightarrow 7{\cos ^2}{\rm{\theta }} + 3 - 3{\cos ^2}{\rm{\theta }} = 4\]
Now by simplifying the above equation, we get
\[ \Rightarrow 4{\cos ^2}{\rm{\theta }} = 1\]
\[ \Rightarrow {\cos ^2}{\rm{\theta }} = \dfrac{1}{4}\]
Now by solving this equation we will get the value of theta. Therefore, we get
\[ \Rightarrow \cos {\rm{\theta }} = \dfrac{1}{2}\]
\[ \Rightarrow {\rm{\theta = 6}}{{\rm{0}}^0}\]
Now we have to put the value of the theta in \[\cot {\rm{\theta }}\] to find the answer. So, we get
\[ \Rightarrow \cot {\rm{\theta }} = \cot {\rm{6}}{{\rm{0}}^0} = \dfrac{1}{{\sqrt 3 }}\]
Hence, \[\dfrac{1}{{\sqrt 3 }}\] is the value of \[\cot {\rm{\theta }}\].
So, option D is the correct option.
Note: We should know the different properties of the trigonometric function and also in which quadrant which function is positive or negative as in the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec is positive. In the second quadrant, only sin and cosec function is positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive. With the help of this concept, this question can be easily solved.
Complete step-by-step answer:
Given equation is \[7{\cos ^2}{\rm{\theta }} + 3{\sin ^2}{\rm{\theta }} = 4\]………… (1)
And we have to find out the value of \[\cot {\rm{\theta }}\].
So, firstly we have to use the basic identity of the trigonometric functions to simplify the equation to find the value of the theta.
Here, we know that \[{\rm{si}}{{\rm{n}}^2}{\rm{\theta }} + {\cos ^2}{\rm{\theta = 1}}\], we can also write it as \[{\rm{si}}{{\rm{n}}^2}{\rm{\theta = 1}} - {\cos ^2}{\rm{\theta }}\] . So, we will put the value of the \[{\sin ^2}{\rm{\theta }}\] in the main equation i.e. equation (1), we get
\[ \Rightarrow 7{\cos ^2}{\rm{\theta }} + 3({\rm{1}} - {\cos ^2}{\rm{\theta )}} = 4\]
\[ \Rightarrow 7{\cos ^2}{\rm{\theta }} + 3 - 3{\cos ^2}{\rm{\theta }} = 4\]
Now by simplifying the above equation, we get
\[ \Rightarrow 4{\cos ^2}{\rm{\theta }} = 1\]
\[ \Rightarrow {\cos ^2}{\rm{\theta }} = \dfrac{1}{4}\]
Now by solving this equation we will get the value of theta. Therefore, we get
\[ \Rightarrow \cos {\rm{\theta }} = \dfrac{1}{2}\]
\[ \Rightarrow {\rm{\theta = 6}}{{\rm{0}}^0}\]
Now we have to put the value of the theta in \[\cot {\rm{\theta }}\] to find the answer. So, we get
\[ \Rightarrow \cot {\rm{\theta }} = \cot {\rm{6}}{{\rm{0}}^0} = \dfrac{1}{{\sqrt 3 }}\]
Hence, \[\dfrac{1}{{\sqrt 3 }}\] is the value of \[\cot {\rm{\theta }}\].
So, option D is the correct option.
Note: We should know the different properties of the trigonometric function and also in which quadrant which function is positive or negative as in the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec is positive. In the second quadrant, only sin and cosec function is positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive. With the help of this concept, this question can be easily solved.
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