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Hint: In this question, we are given the value of the cotangent of the angle and the quadrant in which the angle x lies. Therefore, using the definition of cot(x), and other trigonometric formulas, we can obtain the values of other trigonometric ratios by solving the corresponding equations.

Complete step-by-step answer:

We are given the value of cot(x). We can use the trigonometric relation

$\text{cose}{{\text{c}}^{2}}\left( x \right)=1+{{\cot }^{2}}\left( x \right)$

With the given value of cot(x) to find

$\begin{align}

& \text{cose}{{\text{c}}^{2}}\left( x \right)=1+{{\left( \dfrac{3}{4} \right)}^{2}}=1+\dfrac{9}{16}=\dfrac{16+9}{16}=\dfrac{25}{16} \\

& \Rightarrow \text{cosec}(x)=\pm \sqrt{\dfrac{25}{16}}=\pm \dfrac{5}{4} \\

\end{align}$

However, in the third quadrant, the value of cosec(x) is negative, so we should take only the positive value in the above equation to obtain

$\text{cosec}(x)=\dfrac{-5}{4}............(1.1)$

Also tan(x) and cot(x) are related by

$\tan (x)=\dfrac{1}{\cot (x)}$

Therefore, using the value of cot(x) in the above equation, we find

$\tan (x)=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}................(1.2)$

We know that cosec(x) and sin(x) are related as

$\text{cosec}(x)=\dfrac{1}{\sin (x)}\Rightarrow \sin (x)=\dfrac{1}{\text{cosec}(x)}.........(1.3)$

Therefore, using the value of (1.1) in (1.3), we obtain

$\sin (x)=\dfrac{1}{\text{cosec}(x)}=\dfrac{1}{\dfrac{-5}{4}}=\dfrac{-4}{5}..........(1.4)$

Also, we know that

$\tan (x)=\dfrac{\sin (x)}{\cos (x)}\Rightarrow \cos (x)=\dfrac{\sin (x)}{\tan (x)}$

Therefore, using equations (1.2) and (1.4), we obtain

$\cos (x)=\dfrac{\dfrac{-4}{5}}{\dfrac{4}{3}}=\dfrac{-3}{5}..................(1.5)$

Also, sec(x) is given by

$\sec (x)=\dfrac{1}{\cos (x)}=\dfrac{1}{\dfrac{-3}{5}}=\dfrac{-5}{3}.............(1.6)$

Therefore, from equations (1.1), (1.2), (1.4), (1.5) and (1.5), we have found the other trigonometric ratios as

$\sin (x)=\dfrac{-4}{5}$, $\cos (x)=\dfrac{-3}{5}$, $\tan (x)=\dfrac{4}{3}$, $\sec (x)=\dfrac{-5}{3}$, $\text{cosec}(x)=\dfrac{-5}{4}$

Which is the required answer to this question.

Note: We could also have found tan(x) by using $\tan (x)=\dfrac{1}{\cot (x)}$ and then sec(x) from tan(x) by using the identity ${{\sec }^{2}}\left( x \right)=1+{{\tan }^{2}}\left( x \right)$ and from there we could have found sin(x) and cos(x). However, the answer would have remained the same as found out in the solution above. Also, we should check that as x lies in the third quadrant, both the values of sin(x) and cos(x) are negative.

Complete step-by-step answer:

We are given the value of cot(x). We can use the trigonometric relation

$\text{cose}{{\text{c}}^{2}}\left( x \right)=1+{{\cot }^{2}}\left( x \right)$

With the given value of cot(x) to find

$\begin{align}

& \text{cose}{{\text{c}}^{2}}\left( x \right)=1+{{\left( \dfrac{3}{4} \right)}^{2}}=1+\dfrac{9}{16}=\dfrac{16+9}{16}=\dfrac{25}{16} \\

& \Rightarrow \text{cosec}(x)=\pm \sqrt{\dfrac{25}{16}}=\pm \dfrac{5}{4} \\

\end{align}$

However, in the third quadrant, the value of cosec(x) is negative, so we should take only the positive value in the above equation to obtain

$\text{cosec}(x)=\dfrac{-5}{4}............(1.1)$

Also tan(x) and cot(x) are related by

$\tan (x)=\dfrac{1}{\cot (x)}$

Therefore, using the value of cot(x) in the above equation, we find

$\tan (x)=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}................(1.2)$

We know that cosec(x) and sin(x) are related as

$\text{cosec}(x)=\dfrac{1}{\sin (x)}\Rightarrow \sin (x)=\dfrac{1}{\text{cosec}(x)}.........(1.3)$

Therefore, using the value of (1.1) in (1.3), we obtain

$\sin (x)=\dfrac{1}{\text{cosec}(x)}=\dfrac{1}{\dfrac{-5}{4}}=\dfrac{-4}{5}..........(1.4)$

Also, we know that

$\tan (x)=\dfrac{\sin (x)}{\cos (x)}\Rightarrow \cos (x)=\dfrac{\sin (x)}{\tan (x)}$

Therefore, using equations (1.2) and (1.4), we obtain

$\cos (x)=\dfrac{\dfrac{-4}{5}}{\dfrac{4}{3}}=\dfrac{-3}{5}..................(1.5)$

Also, sec(x) is given by

$\sec (x)=\dfrac{1}{\cos (x)}=\dfrac{1}{\dfrac{-3}{5}}=\dfrac{-5}{3}.............(1.6)$

Therefore, from equations (1.1), (1.2), (1.4), (1.5) and (1.5), we have found the other trigonometric ratios as

$\sin (x)=\dfrac{-4}{5}$, $\cos (x)=\dfrac{-3}{5}$, $\tan (x)=\dfrac{4}{3}$, $\sec (x)=\dfrac{-5}{3}$, $\text{cosec}(x)=\dfrac{-5}{4}$

Which is the required answer to this question.

Note: We could also have found tan(x) by using $\tan (x)=\dfrac{1}{\cot (x)}$ and then sec(x) from tan(x) by using the identity ${{\sec }^{2}}\left( x \right)=1+{{\tan }^{2}}\left( x \right)$ and from there we could have found sin(x) and cos(x). However, the answer would have remained the same as found out in the solution above. Also, we should check that as x lies in the third quadrant, both the values of sin(x) and cos(x) are negative.

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Find the values of other five trigonometric functions if

$\cot (x)=\dfrac{3}{4}$, x lies in the third quadrant,

$\cot (x)=\dfrac{3}{4}$, x lies in the third quadrant,

Trigonometric Functions NCERT EXERCISE 3.2 (Q.3) | Class 11 Maths | Abhishek Sir

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