Question

If the value of \[\tan \theta =\dfrac{8}{15}\] , find all other trigonometric ratios.

Answer 1

Hint: In this question, we are given the value of \[\tan \theta \] and we are asked to find all other trigonometric ratios. We have to find all the trigonometric ratio, step by step. We know the identity, \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] . Using this relation, \[\sec \theta \] can be obtained. And using the value of \[\sec \theta \] , \[\cos \theta \] can be calculated. We also know the identity, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Using this identity, \[\sin \theta \] can be calculated. 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 the question, it is given that, \[\tan \theta =\dfrac{8}{15}\]………………….(1)

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

Taking \[{{\tan }^{2}}\alpha \] 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 θ. Putting the value of tan θ from equation (1) in

equation (4), we get

\[\begin{align}

  & \sec \theta =\sqrt{1+{{\left( \dfrac{8}{15} \right)}^{2}}} \\

 & \Rightarrow \sec \theta =\sqrt{1+\dfrac{64}{225}} \\

 & \Rightarrow \sec \theta =\sqrt{\dfrac{225+64}{225}} \\

 & \Rightarrow \sec \theta =\sqrt{\dfrac{289}{225}} \\

 & \Rightarrow \sec \theta =\dfrac{17}{15} \\

\end{align}\]

Now, we have, \[sec\theta =\dfrac{17}{15}\]……………….(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{17}{15}} \\

 & \Rightarrow \cos \theta =\dfrac{15}{17} \\

\end{align}\]

Now, we also have, \[\cos \theta =\dfrac{15}{17}\]………………..(7)

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

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{15}{17} \right)}^{2}}} \\

 & \Rightarrow sin\theta =\sqrt{1-\dfrac{225}{289}} \\

 & \Rightarrow sin\theta =\sqrt{\dfrac{289-225}{289}} \\

 & \Rightarrow sin\theta =\sqrt{\dfrac{64}{289}} \\

 & \Rightarrow sin\theta =\dfrac{8}{17} \\

\end{align}\]

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{8}{17}, \\

 & \cos ec\theta =\dfrac{1}{\sin \theta }=\dfrac{17}{8}. \\

\end{align}\]

\[\begin{align}

  & \cos \theta =\dfrac{15}{17}, \\

 & sec\theta =\dfrac{1}{cos\theta }=\dfrac{17}{15}. \\

\end{align}\]

\[\begin{align}

  & tan\theta =\dfrac{8}{15}, \\

 & \cot \theta =\dfrac{1}{tan\theta }=\dfrac{15}{8}. \\

\end{align}\]

Now, we have got all the trigonometric ratios.


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

We have, \[\tan \theta =\dfrac{8}{15}\] .

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

Using Pythagora's theorem, we can find the hypotenuse.

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

\[\begin{align}

  & \sqrt{{{\left( 15 \right)}^{2}}+{{(8)}^{2}}} \\

 & =\sqrt{225+64} \\

 & =\sqrt{289} \\

 & =17 \\

\end{align}\]

\[\begin{align}

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

 & \cos \theta =\dfrac{15}{17} \\

\end{align}\]

We know that,

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

\[\sin \theta =\dfrac{8}{17}\]

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