Question

hey mate please solve hey mate solve it please .​

Answer 1

SOLUTION:-

Given:

(1+4x²)cosA = 4x

Therefore,

$secx = \frac{1 + 4 {x}^{2} }{4x} = \frac{Hypotenuse}{Base}$

So,

Using Pythagoras rule;

$Perpendicular = \sqrt{(1 + 4 {x}^{2} ) {}^{2} - (16{x}^{2} } ) {} \\ \\ = > \sqrt{1 + 16 {x}^{4} + 8 {x}^{2} - 16 {x}^{2} } \\ \\ = > \sqrt{(1 - 4 {x}^{2} ) {}^{2} } \\ \\ = > 1 - 4 {x}^{2}$

Now,

$cosecA - cotA = \frac{1 - 4 {x}^{2} }{1 + 4 {x}^{2} } - \frac{1 - 4 {x}^{2} }{4x} \\ \\ = > \frac{(1 - 2x)(1 + 2x)}{(1 + 2x) {}^{2} } \\ \\ = > \frac{(1 - 2x)}{(1 + 2x)}$

Proved.

Hope it helps ☺️

Answer 2

Step-by-step explanation:

Hola mate!!!!

Given that

(1+4x^2) Cos A = 4x

$\cos(a) = \frac{4x}{(1 + 4 {x}^{2}) }$

$we \: know \: that = > \\ \cos(a) = \frac{base}{hypo}$

$now \: comparing \: it \: we \: will \: get$

$base = 4x \: and \: hypo = (1 + 4 {x}^{2} )$

$so \: perpendicular \: will \: be \: (1 - 4 {x}^{2} )$

$\csc(a) = \frac{hypo}{perpendicular}$

$\csc(a) = \frac{(1 + 4 {x}^{2}) }{(1 - 4 {x}^{2} )}$

$\cot(a) = \frac{base}{perpendicular}$

$\cot(a) = \frac{4x}{(1 - 4 {x}^{2} )}$

$\csc(a) + \cot( a) \\ = \frac{(1 + 4 {x}^{2}) }{(1 - 4 {x}^{2} )} + \frac{4x}{(1 - 4 {x}^{2} )}$

$= \frac{1 + 4 {x}^{2} + 4x }{(1 - 4 {x}^{2} )}$

$= \frac{ {1 + 2x}^{2} }{(1 - 2x)(1 + 2x)}$

$= \frac{(1 + 2x)}{(1 - 2x)}$

$hope \: it \: helps \: uh.....$