Value of $tan 225^{o}$ is :

Open in App

Solution

$tan 225^{o}$ value will be $+$ ve
as it lie in $3^{r d}$ quadrant
and $tan θ$ is positive in third quadrant.
$tan (180 - 45^{o})$
$\Rightarrow tan 45^{o}$
$= 1$ .

Suggest Corrections

Similar questions

\frac{cot 81^{o} cot 69^{o} - tan 225^{o}}{cot 261^{o} + tan 21^{o}}

\frac{{sin}^{2} θ + {sin}^{2} (90^{o} - θ)}{3 ({sec}^{2} 61^{o} - {cot}^{2} 29^{o})} - \frac{3 {cot}^{2} 30^{o} . {sin}^{2} 54^{o} . {sec}^{2} 36^{o}}{2 ({csc}^{2} 65^{o} - {tan}^{2} 25 o)}

A B C

y

x

z

y

x

| z - 2 + 2 i | = 1,

Join BYJU'S Learning Program