Question

Hi there Cotx-Tanx = please give proper solutions with explanations then I will mark as brainliest no spamming

Answer 1

Answer:

we know that cotθ is

$\frac{1}{tanθ}$

Step-by-step explanation:

so we can write cotx - tanx as :

$\frac{1}{tanx} \: - tanx$

$\frac{1 - tan {}^{2}x }{tan x\: }$

(LCM taken)

We know that tanθ =

$\frac{p}{b}$

where P = perpendicular

B = base

∴

$\frac{1 - \frac{p {}^{2} }{b {}^{2} } }{ \frac{p}{b} }$

$\frac{b {}^{2} - p {}^{2} }{b {}^{2} } \div \frac{p}{b}$

$\frac{ {b}^{2} - \: {p}^{2} }{ {b}^{2} } \times \frac{b}{p}$

$\frac{ {b}^{2} - \: {p \:}^{2} }{bp}$

then using a² - b² = (a+b)(a-b)

$\frac{ \: (b+p)(b-p) \: }{pb}$

identity

we can get the simplest form :