Question

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

Answer 1

We have $\sin A=\frac{9}{41}$,

As,

$$\begin{gathered} {\cos }^{2}A=1-{\sin }^{2}A \\ =1-{\left(\frac{9}{41}\right)}^2 \\ =1-\frac{81}{1681} \\ =\frac{1681-81}{1681} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\cos }^2\mathrm{A}=\frac{1600}{1681} \\ \Rightarrow \cos A=\sqrt{\frac{1600}{1681}} \\ \Rightarrow \cos A=\frac{40}{41} \end{gathered}$$

Also,

$$\begin{gathered} \tan A=\frac{\sin A}{\cos A} \\ =\frac{\left({\displaystyle \frac{9}{41}}\right)}{\left({\displaystyle \frac{40}{41}}\right)} \\ =\frac{9}{40} \end{gathered}$$