If (( 1−sin(x)−cos(x))/(1+sin(x)−cos(x))) = (1/4) then find the value of: tan(x) + (1/(cos(x))) =?


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Question Number 161931 by mnjuly1970 last updated on 24/Dec/21

$I f \frac{1 - s i n (x) - c o s (x)}{1 + s i n (x) - c o s (x)} = \frac{1}{4}$ $t h e n f i n d t h e v a l u e o f :$ $t a n (x) + \frac{1}{c o s (x)} = ?$
Commented by cortano last updated on 24/Dec/21
$((2sin^2 ((x/2))−2sin ((x/2))cos ((x/2)))/(2sin^2 ((x/2))+2sin ((x/2))cos ((x/2)))) =(1/4) ((sin (x/2)−cos (x/2))/(sin (x/2)+cos (x/2))) =(1/4) ((1−sin x)/(cos x)) = (1/4) sec x−tan x=(1/4) . let tan x+sec x =p { ((sec^2 x−tan^2 x=(p/4))),((tan^2 x+1−tan^2 x=(p/4)⇒p=4)) :} tan x+sec x = 4$
$\frac{2 \sin^{2} (\frac{x}{2}) - 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})}{2 \sin^{2} (\frac{x}{2}) + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})} = \frac{1}{4}$ $\frac{\sin \frac{x}{2} - \cos \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} = \frac{1}{4}$ $\frac{1 - \sin x}{\cos x} = \frac{1}{4}$ $\sec x - \tan x = \frac{1}{4} . l e t \tan x + \sec x = p$ ${\begin{cases} \sec^{2} x - \tan^{2} x = \frac{p}{4} \\ \tan^{2} x + 1 - \tan^{2} x = \frac{p}{4} \Rightarrow p = 4 \end{cases}$ $\tan x + \sec x = 4$
Commented by cortano last updated on 24/Dec/21

$a n o t h e r w a y$ $\frac{(1 - \sin x) (1 + \sin x)}{\cos x (1 + \sin x)} = \frac{1}{4}$ $\frac{\cos x}{1 + \sin x} = \frac{1}{4} \Rightarrow \frac{1 + \sin x}{\cos x} = 4$ $\Leftrightarrow \frac{1}{\cos x} + \tan x = 4$
Commented by peter frank last updated on 25/Dec/21

$great$
Commented by mnjuly1970 last updated on 24/Dec/21

$v e r y n i c e s o l u t i o n s i r . . . t h a n k y o u$
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