Question

If : √3 cot2Θ - 4 cot Θ + √3 = 0, then find the value of ( cot2Θ + tan2 Θ)

Answer 1

Answer:

ye kon c class ka q hai ?

Answer 2

Answer:

$\frac{10}{3}$

Step-by-step explanation:

$\sqrt{3}$ cot² θ - 4 cot θ + $\sqrt{3}$ = 0

Let cot θ = x

then $\sqrt{3} x^{2} - 4x + \sqrt{3} = 0$

which is a polynomial of the form $ax^{2} + bx + c = 0$

=> a = $\sqrt{3$, b = -4, c = $\sqrt{3$

=> x = $\frac{-b}{2a}$±$\frac{\sqrt{b^{2} - 4ac}}{2a}$

=> x = $\frac{-(-4) +- \sqrt{(-4)^{2} - 4(\sqrt{3})(\sqrt{3})}}{2(\sqrt{3})}$     (+- denotes ±) (Quadratic Formula)

=> x = $\frac{4+- \sqrt{16 - 12}}{2\sqrt{3}}$

=> x = $\frac{4+- \sqrt{4}}{2\sqrt{3}}$

=> x = $\frac{4+- 2}{2\sqrt{3}}$

=> x = $\frac{2+- 1}{\sqrt{3}}$

=> x = $\frac{1}{\sqrt{3}}$, $\sqrt{3}$

Then cot² θ + tan² θ:

= $(\frac{1}{\sqrt{3}} )^{2}$ + $(\sqrt{3})^{2}$ (or) $(\sqrt{3})^{2} + (\frac{1}{\sqrt{3}})^{2}$

= $\frac{1}{3} + 3$ (or) $3 + \frac{1}{3}$

= $\frac{10}{3}$