
Given \[\sin x=\dfrac{4}{7}\] and \[\cos x=-\dfrac{\sqrt{33}}{7}\], how do you find cot x?
Answer
532.5k+ views
Hint: This type of problem is based on the concept of trigonometry. We have been given values of sin x and cos x. We know that the expansion of cot x is \[\cot x=\dfrac{\cos x}{\sin x}\]. Substitute the values of sin x and cos x. the, using the property of division that is \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\], we can further simplify the given equation. Here, we get a= \[-\sqrt{33}\], b=7, c=4 and d=7. Cancel 7 from the numerator and denominator. Do necessary calculations and find the value of x.
Complete step by step solution:
According to the question, we are asked to find the value of cot x.
We have been given the values of \[\sin x=\dfrac{4}{7}\] ------------(1)
And \[\cos x=-\dfrac{\sqrt{33}}{7}\]
We can express the value of cos x as \[\cos x=\dfrac{-\sqrt{33}}{7}\]. ----------(2)
We know that the expansion of cot x is cos x divided by sin x.
That is \[\cot x=\dfrac{\cos x}{\sin x}\].
Let us substitute the values from (1) and (2) to the expansion.
\[\Rightarrow \cot x=\dfrac{\dfrac{-\sqrt{33}}{7}}{\dfrac{4}{7}}\]
We know that \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\]. Let us use this property of division in the above expression.
Here, we get a= \[-\sqrt{33}\], b=7, c=4 and d=7.
\[\Rightarrow \cot x=\dfrac{-\sqrt{33}}{7}\times \dfrac{7}{4}\]
We find that 7 are common in both the numerator and denominator. On cancelling 7 from the numerator and denominator, we get
\[\cot x=\dfrac{-\sqrt{33}}{1}\times \dfrac{1}{4}\]
Therefore, we get \[\cot x=\dfrac{-\sqrt{33}}{4}\].
Hence, the value of cot x when \[\sin x=\dfrac{4}{7}\] and \[\cos x=-\dfrac{\sqrt{33}}{7}\] is \[\dfrac{-\sqrt{33}}{4}\].
Note: We can also solve this problem by first finding the value of tan x by dividing the value of sin x by cos x and then taking the reciprocal of tan x, that is \[\cot x=\dfrac{1}{\tan x}\]. This method will have more steps. We have to cancel all the common terms from the fraction. Don’t get confused by the expansion of cot x and tan x. The expansion of tan x is \[\tan x=\dfrac{\sin x}{\cos x}\] and the expansion of cot x is \[\cot x=\dfrac{\cos x}{\sin x}\].
Complete step by step solution:
According to the question, we are asked to find the value of cot x.
We have been given the values of \[\sin x=\dfrac{4}{7}\] ------------(1)
And \[\cos x=-\dfrac{\sqrt{33}}{7}\]
We can express the value of cos x as \[\cos x=\dfrac{-\sqrt{33}}{7}\]. ----------(2)
We know that the expansion of cot x is cos x divided by sin x.
That is \[\cot x=\dfrac{\cos x}{\sin x}\].
Let us substitute the values from (1) and (2) to the expansion.
\[\Rightarrow \cot x=\dfrac{\dfrac{-\sqrt{33}}{7}}{\dfrac{4}{7}}\]
We know that \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\]. Let us use this property of division in the above expression.
Here, we get a= \[-\sqrt{33}\], b=7, c=4 and d=7.
\[\Rightarrow \cot x=\dfrac{-\sqrt{33}}{7}\times \dfrac{7}{4}\]
We find that 7 are common in both the numerator and denominator. On cancelling 7 from the numerator and denominator, we get
\[\cot x=\dfrac{-\sqrt{33}}{1}\times \dfrac{1}{4}\]
Therefore, we get \[\cot x=\dfrac{-\sqrt{33}}{4}\].
Hence, the value of cot x when \[\sin x=\dfrac{4}{7}\] and \[\cos x=-\dfrac{\sqrt{33}}{7}\] is \[\dfrac{-\sqrt{33}}{4}\].
Note: We can also solve this problem by first finding the value of tan x by dividing the value of sin x by cos x and then taking the reciprocal of tan x, that is \[\cot x=\dfrac{1}{\tan x}\]. This method will have more steps. We have to cancel all the common terms from the fraction. Don’t get confused by the expansion of cot x and tan x. The expansion of tan x is \[\tan x=\dfrac{\sin x}{\cos x}\] and the expansion of cot x is \[\cot x=\dfrac{\cos x}{\sin x}\].
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