Question

If 2 cos square theta - 5 cos theta + 2 is equal to zero then the value of cos theta + cot theta

Answer 1

Answer:

$\underline{\bigstar\:\:\textsf{First Part of the Question :}}$

$:\implies\sf 2\cos^2(\theta)-5\cos(\theta)+2=0\\\\\\:\implies\sf 2\cos^2(\theta)-(4+1)\cos(\theta)+2=0\\\\\\:\implies\sf 2\cos^2(\theta)-4\cos(\theta)-\cos(\theta)+2=0\\\\\\:\implies\sf 2\cos(\theta)(\cos(\theta)-2)-1(\cos(\theta)-2)=0\\\\\\:\implies\sf (2\cos(\theta)-1)(\cos(\theta)-2)=0\\\\\\:\implies\sf \cos(\theta)=\dfrac{1}{2}\quad or\quad \cos(\theta)=2\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{we will take \tt\frac{1}{2} , as range of \cos is from -1 to 1.}}\\\\:\implies\sf \cos(\theta)=\dfrac{1}{2}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{\cos\:60^{\circ}=\frac{1}{2}}}\\\\:\implies\sf \cos(\theta)=\cos(60^{\circ})\\\\\\:\implies\underline{\boxed{\sf\theta=60^{\circ}}}$

$\rule{150}{1}$

$\underline{\bigstar\:\:\textsf{According to the Question :}}$

$\dashrightarrow\sf\:\:\cos(\theta)+\cot(\theta)\\\\\\\dashrightarrow\sf\:\:\cos(60^{\circ})+\cot(60^{\circ})\\\\\\\dashrightarrow\sf\:\:\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\\\\\\\dashrightarrow\sf\:\:\dfrac{1}{2} +\dfrac{\sqrt{3}}{3}\\\\\\\dashrightarrow\:\:\underline{\boxed{\textsf{\textbf{ \dfrac{ \text3\:\text+\:\text2\sqrt{\text3}}{\text6} }}}}$

Answer 2

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✺ Answer :---

$\implies\ \ {2\cos^2\theta-5\cos\theta+2=0}$

$\implies\ \ {2\cos^2\theta-(4+1)\cos\theta+2=0}$

$\implies\ \ {2\cos^2\theta-4\cos\theta-\cos\theta+2=0}$

$\implies\ \ {2\cos\theta(\cos\theta-2)-1(\cos\theta-2)=0}$

$\implies\ \ {(\cos\theta-2)(2\cos\theta-1)=0}$

So , either ,

$\blacksquare\:\:\:\ \ {(\cos\theta-2)=0}$

$\longrightarrow\ \ {\cos\theta=2}$

Or,

$\blacksquare\:\:\:\ \ {(2\cos\theta-1)=0}$

$\longrightarrow\ \ {2\cos\theta=1}$

$\longrightarrow\ \ {</p><p>\cos\theta=\dfrac{1}{2}}$

Now $\small{\text{from the definition of} \cos\theta \text{ we know that,}}$

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$\blacksquare\:\:\:\red{ - 1 \leqslant \cos \theta \leqslant + 1}$

✏ so we will take the value of ,

$\longrightarrow\ \ {</p><p>\cos\theta=\dfrac{1}{2}}$

$\longrightarrow\ \ {</p><p>\cos\theta=\cos60\degree}$

$\longrightarrow\ \ {\boxed{\theta=60\degree}}$

∴ $\ \ {\cos\theta+\cot\theta}$

$\implies\ \ {\cos60\degree+\cot60\degree}$

$\implies\ \ {\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}}$

$\implies\ \ {\dfrac{\sqrt{3}+2}{2\sqrt{3}}}$

$\implies\ \ {\large{\dfrac{2+\sqrt{3}}{2\sqrt{3}}}}$

$\implies\ \ {\large{\dfrac{(2+\sqrt{3})\times\sqrt{3}}{2\sqrt{3}\times\sqrt{3}}}}$

$\implies\ \ {\bf{\large{\dfrac{2\sqrt{3}+3}{6}}}}$

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