If-1-sin-x-cos-x-1-sin-x-cos-x-1-4-then-find-the-value-of-tan-x-1-cos-x-

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

\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 \dots t h a n k y o u