Question

If $\sin A=\dfrac{9}{41}$ ,find the value of cosA and tanA.

Answer 1

Hint: In this question, from the given values of sin function by using the trigonometric identities we can find the values of the tan and cos functions. Then on substituting the respective values in the given expression of the question we can calculate the left hand side value and the right hand side value.

Complete step-by-step answer:
Then on comparing the values obtained, we get the result.

& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\\ \end{aligned}$$ Now, from the given question we have $$\sin A=\dfrac{9}{41}\ \ \ \ \ ...(a)$$ Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following substitute the value from the question and as well as from equation (a) $$\begin{aligned} & \Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\\ & \Rightarrow {{\cos }^{2}}A=1-{{\left( \dfrac{9}{41} \right)}^{2}} \\\ & \Rightarrow {{\cos }^{2}}A=1-\dfrac{81}{1681} \\\ & \Rightarrow \cos A=\sqrt{\dfrac{1681-81}{1681}}=\sqrt{\dfrac{1600}{1681}} \\\ & \Rightarrow \cos A=\dfrac{40}{41} \\\ \end{aligned}$$ Now, again from the hint, we know that we can get the tan function as follows $$\begin{aligned} & \Rightarrow \tan A=\dfrac{\sin A}{\cos A} \\\ & \Rightarrow \tan A=\dfrac{\dfrac{9}{41}}{\dfrac{40}{41}} \\\ & \Rightarrow \tan A=\dfrac{9}{40} \\\ \end{aligned}$$ Hence, we found the required values as follows $$\begin{aligned} & \cos A=\dfrac{40}{41} \\\ & \tan A=\dfrac{9}{40} \\\ \end{aligned}$$ Note: In such kind of questions, one must be well versed with the relations of trigonometric functions because, without them one could never get to the correct solution. It is important to note that while calculating the values of respective functions we need to use the identities accordingly and solve them. Because neglecting any of the terms or writing it incorrectly changes the complete result.