Question

If $\sin \theta =\frac{40}{41}$, then $\tan \theta =......$ and $\cos \theta =.....$

• A. $\frac{40}{9},\frac{9}{41}$
• B. $\frac{40}{41},\frac{9}{41}$
• C. $\frac{41}{9},\frac{41}{40}$
• D. None of these

Answer 1

The correct option is A $\frac{40}{9},\frac{9}{41}$

Using Pythagorean identity

${\sin }^2\theta +{\cos }^2\theta =1$
${\left(\frac{40}{41}\right)}^2+{\cos }^2\theta =1$
So, ${\cos }^2\theta =1-\frac{1600}{1681}=\frac{81}{1681}$
So, $\cos \theta =\frac{9}{41}$
and $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\frac{40}{41}}{\frac{9}{41}}$
$=\frac{40\times 41}{9\times 41}=\frac{40}{9}$