Question

Find the value of $x$ in the following:
$\tan x=\sin {45}^{\circ }\cos {45}^{\circ }+\sin {30}^{\circ }$

Answer 1

Given $\tan x=\sin {45}^{\circ }\cos {45}^{\circ }+\sin {30}^{\circ }$

$\tan x=(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})+\frac{1}{2}$ $[\because \sin {30}^{\circ }=\frac{1}{2},\cos {45}^{\circ }=\sin {45}^{\circ }=\frac{1}{\sqrt{2}}]$

$\tan x={\left(\frac{1}{\sqrt{2}}\right)}^2+\frac{1}{2}$

$\tan x=\frac{1}{2}+\frac{1}{2}$

$\tan x=\frac{2}{2}$

$\tan x=1$

$\tan {45}^{\circ }=1$

$\tan x=\tan {45}^{\circ }$

$\therefore x={45}^{\circ }$