Question

If the value of \[cot\theta =\dfrac{12}{5}\] , find all other trigonometric ratios.

Answer 1

Hint: In this question, we are given the value of \[\cot \theta \] . With the help of \[\cot \theta \] we can find \[\tan \theta \] . Here, we are asked to find all other trigonometric ratios. We have to find all the trigonometric ratios, step by step. We know the relation between \[\tan \theta \] and \[\sec \theta \] . Using this relation, \[\sec \theta \] can be obtained. And, using the value of \[\sec \theta \] , \[\cos \theta \] can be obtained. Also, using the relation between \[\cos \theta \] and \[\sin \theta \] , \[\sin \theta \] can be obtained. If we have got the values of \[\sin \theta \], \[\cos \theta \] and \[\tan \theta \] , then we can find all other trigonometric ratios easily.

Complete step-by-step answer:
Now, according to the question, it is given that

\[cot\theta =\dfrac{12}{5}\]

\[\tan \theta =\dfrac{5}{12}\]………………….(1)

We know that, \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]……………………(2)

Taking tan2 θ to the RHS in the equation (2), we get

\[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]…………….(3)

Now, \[\sec \theta \] can be easily expressed in terms of \[\tan \theta \] .

Taking square root in both LHS and RHS in equation (3), we get

\[\sec \theta =\sqrt{1+{{\tan }^{2}}\theta }\]……………….(4)

In question, we are given the value of \[\tan \theta \] . Putting the value of \[\tan \theta \]

from equation (1) in equation (4), we get

\[\begin{align}

  & \sec \theta =\sqrt{1+{{\left( \dfrac{5}{12} \right)}^{2}}} \\

 & \Rightarrow \sec \theta =\sqrt{1+\dfrac{25}{144}} \\

 & \Rightarrow \sec \theta =\sqrt{\dfrac{144+25}{144}} \\

 & \Rightarrow \sec \theta =\sqrt{\dfrac{169}{144}} \\

 & \Rightarrow \sec \theta =\dfrac{13}{12} \\

\end{align}\]

Now, we have, \[sec\theta =\dfrac{13}{12}\]……………….(5)

From equation (1) and equation (5), we have got the values of \[\sec \theta \] and \[\tan

\theta \] .

Using equation (5), we can find the value of \[\cos \theta \] .

We know that, \[\dfrac{1}{sec\theta }=\cos \theta\]…………………….(6)

Putting the values of \[\sec \theta \] in equation (6), we get

\[\begin{align}

  & \cos \theta =\dfrac{1}{\sec \theta } \\

 & \Rightarrow \cos \theta =\dfrac{1}{\dfrac{13}{12}} \\

 & \Rightarrow \cos \theta =\dfrac{12}{13} \\

\end{align}\]

Now, we also have

\[\cos \theta =\dfrac{12}{13}\]………………..(7)

We have to find other remaining trigonometric ratios that are \[\sin \theta \],

\[\operatorname{cosec}\theta \] , and \[\cot \theta \] .

We know the identity,

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]……………………(8)

Taking \[{{\sin }^{2}}\theta \] to the RHS in the equation (8), we get

\[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]…………….(9)

Now, \[\sin \theta \] can be easily expressed in terms of \[\cos \theta \] .

Now, \[\sin \theta \] can be easily expressed in terms of \[\cos \theta \] .

Taking square root in both LHS and RHS in equation (9), we get

\[sin\theta =\sqrt{1-{{\cos }^{2}}\theta }\]……………(10)

In equation(7), we have got the value of \[\cos \theta \] . Putting the value of \[\cos \theta \]

from equation (7) in equation (10), we get

\[\begin{align}

  & sin\theta =\sqrt{1-co{{s}^{2}}\theta } \\

 & \Rightarrow sin\theta =\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}} \\

 & \Rightarrow sin\theta =\sqrt{1-\dfrac{144}{169}} \\

 & \Rightarrow sin\theta =\sqrt{\dfrac{169-144}{169}} \\

 & \Rightarrow sin\theta =\sqrt{\dfrac{25}{169}} \\

 & \Rightarrow sin\theta =\dfrac{5}{13} \\

\end{align}\]

We know that \[\sec \theta \] , \[\cos ec\theta \] , and \[\cot \theta \] is reciprocal of \[\cos

\theta \], \[\sin \theta \] and \[\tan \theta \]respectively.

\[\begin{align}

  & \sin \theta =\dfrac{5}{13}, \\

 & \cos ec\theta =\dfrac{1}{\sin \theta }=\dfrac{13}{5}. \\

\end{align}\]

\[\begin{align}

  & \cos \theta =\dfrac{12}{13}, \\

 & sec\theta =\dfrac{1}{cos\theta }=\dfrac{13}{12}. \\

\end{align}\]

\[\begin{align}

  & tan\theta =\dfrac{5}{12}, \\

 & \cot \theta =\dfrac{1}{tan\theta }=\dfrac{12}{5}. \\

\end{align}\]

Now, we have got all the trigonometric ratios.


Note: This question can also be solved by using the Pythagoras theorem.

We have, \[\cot \theta =\dfrac{12}{5}\] .

Now, using a right-angled triangle, we can get the value of \[\cos \theta \].

Using Pythagoras theorem, we can find the hypotenuse.

Hypotenuse = \[\sqrt{{{\left( base \right)}^{2}}+{{\left( height \right)}^{2}}}\]

\[\begin{align}

  & \sqrt{{{\left( 12 \right)}^{2}}+{{(5)}^{2}}} \\

 & =\sqrt{144+25} \\

 & =\sqrt{169} \\

 & =13 \\

\end{align}\]

\[\begin{align}

  & \cos \theta =\dfrac{base}{hypotenuse} \\

 & \cos \theta =\dfrac{12}{13} \\

\end{align}\]

We know that,

\[\sin \theta =\dfrac{height}{hypotenuse}\]

\[\sin \theta =\dfrac{5}{13}\]

Now, other remaining trigonometric ratios can be calculated by finding the reciprocal of these trigonometric ratios.