Question

If the value of \[\cos \theta =\dfrac{7}{25}\], find the value of other trigonometric ratios.

Answer 1

Hint: In this question, we are given the value of \[cos\theta \] 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, \[\sin \theta \] 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, \[\cos \theta =\dfrac{7}{25}\]………………….(1)

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

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

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

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

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

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

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

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

\[\begin{align}

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

 & \Rightarrow sin\theta =\sqrt{1-{{\left( \dfrac{7}{25} \right)}^{2}}} \\

 & \Rightarrow sin\theta =\sqrt{1-\dfrac{49}{625}} \\

 & \Rightarrow sin\theta =\sqrt{\dfrac{625-49}{625}} \\

 & \Rightarrow sin\theta =\sqrt{\dfrac{576}{625}} \\

 & \Rightarrow sin\theta =\dfrac{24}{25} \\

\end{align}\]

Now, we have \[sin\theta =\dfrac{24}{25}\]……………….(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 \[cos\theta \] and \[\sin \theta \] in equation (6), we get

\[\begin{align}

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

 & \Rightarrow \tan \theta =\dfrac{\dfrac{24}{25}}{\dfrac{7}{25}} \\

 & \Rightarrow \tan \theta =\dfrac{24}{7} \\

\end{align}\]

Now, we also have

\[\tan \theta =\dfrac{24}{7}\]………………..(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{24}{25}, \\

 & \cos ec\theta =\dfrac{1}{\sin \theta }=\dfrac{25}{24}. \\

\end{align}\]

\[\begin{align}

  & \cos \theta =\dfrac{7}{25}, \\

 & sec\theta =\dfrac{1}{cos\theta }=\dfrac{25}{7}. \\

\end{align}\]

\[\begin{align}

  & tan\theta =\dfrac{24}{7}, \\

 & \cot \theta =\dfrac{1}{tan\theta }=\dfrac{7}{24}. \\

\end{align}\]

Now, we have got all the trigonometric ratios.

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

We have, \[\cos \theta =\dfrac{7}{25}\] .

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

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

Height = \[\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}\]

\[\begin{align}

  & \sqrt{{{\left( 25 \right)}^{2}}-{{7}^{2}}} \\

 & =\sqrt{625-49} \\

 & =\sqrt{576} \\

 & =24 \\

\end{align}\]

\[\begin{align}

  & sin\theta =\dfrac{height}{hypotenuse} \\

 & sin\theta =\dfrac{24}{25} \\

\end{align}\]

We know that,

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

\[\tan \theta =\dfrac{24}{7}\]

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