Question

If $\cos A=\frac{9}{41}$, find $\tan A$ and $\csc A$

Answer 1

$\cos A=\frac{9}{41}$

$\sec A=\frac{1}{\cos A}=\frac{1}{\frac{9}{41}}=\frac{41}{9}$

We know,

${\sec }^{2}A=1+{\tan }^{2}A$

$\Rightarrow$ $\tan A=\sqrt{{\sec }^{2}A-1}$

$\Rightarrow$ $\tan A=\sqrt{{\left(\frac{41}{9}\right)}^2-1}$

$\Rightarrow$ $\tan A=\sqrt{\frac{1681}{81}-1}$

$\Rightarrow$ $\tan A=\sqrt{\frac{1600}{81}}$

$\Rightarrow$ $\tan A=\frac{40}{9}$

We know,

${\csc }^{2}A=1+{\cot }^{2}A$

$\Rightarrow$ $\csc A=\sqrt{1+{\cot }^{2}A}$

$\Rightarrow$ $\csc A=\sqrt{1+{\left(\frac{9}{40}\right)}^2}$ [ Since, $\cot \theta =\frac{1}{\tan \theta }$ ]

$\Rightarrow$ $\csc A=\sqrt{1+\frac{81}{1600}}$

$\Rightarrow$ $\csc A=\sqrt{\frac{1681}{1600}}$

$\Rightarrow$ $\csc A=\frac{41}{40}$