Question

If the value of \[sin\text{ }\theta =\dfrac{11}{15}\] , find the value of other trigonometric ratios.

Answer 1

Hint: In this question, we are given the value of \[\sin \theta \] and we are asked to find all other trigonometric ratios. We have to find all the trigonometric ratios, step by step. We know the identity, \[{{\sin }^{2}}\theta +\text{ co}{{\text{s}}^{2}}\theta =1\] . Using this relation, \[\cos \theta \] can be obtained. Now, we have the value of \[\sin \theta \] and \[\cos \theta \] . Using the value of \[\sin \theta \] and \[\cos \theta \], \[\tan \theta \] can be calculated. Now, other remaining trigonometric ratios can be calculated by finding the reciprocal of these trigonometric ratios.

Complete step-by-step answer:
Now, according to question, it is given that \[sin\text{ }\theta =\dfrac{11}{15}\]………………….(1)

We know the identity, \[{{\sin }^{2}}\theta +\text{ co}{{\text{s}}^{2}}\theta =1\]……………………(2)

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

\[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \] …………….(3)

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

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

\[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\]……………(4)

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

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

\[\begin{align}

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

 & \Rightarrow \cos \theta =\sqrt{1-{{\left( \dfrac{11}{15} \right)}^{2}}} \\

 & \Rightarrow \cos \theta =\sqrt{1-\dfrac{121}{225}} \\

 & \Rightarrow \cos \theta =\sqrt{\dfrac{225-121}{225}} \\

 & \Rightarrow \cos \theta =\sqrt{\dfrac{104}{225}} \\

 & \Rightarrow \cos \theta =\dfrac{\sqrt{104}}{15} \\

\end{align}\]

Now, we have, \[\cos \theta =\dfrac{\sqrt{104}}{15}\]……………….(5)

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

\theta \] .

Using equation (1) and equation (5), we can find the value of \[\tan \theta \] .

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

Putting the values of cos θ and sin θ in equation (6), we get

\[\begin{align}

  & \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\

 & \Rightarrow \tan \theta =\dfrac{\dfrac{11}{15}}{\dfrac{\sqrt{104}}{15}} \\

 & \Rightarrow \tan \theta =\dfrac{11}{\sqrt{104}} \\

\end{align}\]

Now, we also have \[\tan \theta =\dfrac{11}{\sqrt{104}}\]………………..(7)

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

\[\operatorname{cosec}\theta \] ,

and \[\cot \theta \] . We know that \[\sec \theta \] , \[\operatorname{cosec}\theta \], and

\[\cot \theta \] is reciprocal of \[\cos \theta \], \[\sin \theta \] and \[\tan \theta

\]respectively.

\[\begin{align}

  & \sin \theta =\dfrac{11}{15}, \\

 & \operatorname{cosec}\theta =\dfrac{1}{\sin \theta }=\dfrac{15}{11}. \\

\end{align}\]

\[\begin{align}

  & \cos \theta =\dfrac{\sqrt{104}}{15}, \\

 & sec\theta =\dfrac{1}{cos\theta }=\dfrac{15}{\sqrt{104}}. \\

\end{align}\]

\[\begin{align}

  & tan\theta =\dfrac{11}{\sqrt{104}}, \\

 & \cot \theta =\dfrac{1}{tan\theta }=\dfrac{\sqrt{104}}{11}. \\

\end{align}\]

Now, we have got all the trigonometric ratios.


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

We have, \[\sin \theta =\dfrac{11}{15}\] .


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

Using Pythagoras theorem, we can find the base.

Base = \[\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( height \right)}^{2}}}\]

\[\begin{align}

  & \sqrt{{{\left( 15 \right)}^{2}}-{{11}^{2}}} \\

 & =\sqrt{225-121} \\

 & =\sqrt{104} \\

\end{align}\]

\[\begin{align}

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

 & \cos \theta =\dfrac{\sqrt{104}}{15} \\

\end{align}\]

We know that,

\[\tan \theta =\dfrac{height}{base}\]

\[\tan \theta =\dfrac{11}{\sqrt{104}}\]

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