Question

$cosx+sinx=\frac{1}{2}$ then tan x = ?

Answer 1

We have 0 < x < $\pi$, So from first quadrant, we get minimum value of $cosx+sinx$ is 1, as in first quadrant we know $0<x<\frac{\pi }{2}$
And
We have $cosx+sinx=\frac{1}{2}$
Taking whole square on both hand side, we get
$(cosx+sinx{)}^2=\frac{1}{4}\Rightarrow si{n}^{2}x+co{s}^{2}x+2sinxcosx=\frac{1}{4}\Rightarrow 1+sin2x=\frac{1}{4}(Asweknowsi{n}^{2}x+co{s}^{2}x=1and2sinxcosx=sin2x)\Rightarrow sin2x=\frac{1}{4}-1\Rightarrow sin2x=\frac{-3}{4}$
$si{n}^{2}2x+co{s}^{2}2x=1$, Substitute value, we get
$\Rightarrow {(-\frac{3}{4})}^2+co{s}^{2}2x=1\Rightarrow co{s}^{2}2x=1-\frac{9}{16}\Rightarrow co{s}^{2}2x=\frac{7}{16}\Rightarrow cos2x=\sqrt{\frac{7}{16}}\Rightarrow cos2x=\frac{\sqrt{7}}{4}$
And we know $sin2x=\frac{2tanx}{1+ta{n}^{2}x}Andcos2x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}$, so we get
$tanx=\frac{sin2x}{1+cos2x}$, Substitute values, we get
$\Rightarrow tanx=\frac{-\frac{3}{4}}{1+\frac{\sqrt{7}}{4}}\Rightarrow tanx=\frac{-\frac{3}{4}}{\frac{4+\sqrt{7}}{4}}\Rightarrow tanx=\frac{-3}{4+\sqrt{7}}\Rightarrow tanx=\frac{-3}{4+\sqrt{7}}\times \frac{4-\sqrt{7}}{4-\sqrt{7}}\Rightarrow tanx=\frac{-12+3\sqrt{7}}{16-7}\Rightarrow tanx=\frac{-3(4-\sqrt{7})}{9}\Rightarrow tanx=-\frac{(4-\sqrt{7})}{3}$