How do you verify $(\frac{tan (x)}{1 + sec (x)}) + (1 + \frac{sec (x)}{tan (x)}) = 2 csc (x)$ $(tan(x)/(1 + sec(x))) + (1+sec(x)/tan(x)) = 2csc(x)$ ?

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Answer 1 · 2016-04-16T22:37:23

See below

LHS=left hand side, RHS=right hand side

LHS $= \frac{tan x}{1 + sec x} + \frac{1 + sec x}{tan x}$ $=(tanx)/(1+secx) +(1+secx)/tanx$

$= \frac{(tan x) (tan x) + (1 + sec x) (1 + sec x)}{tan x (1 + sec x)}$ $=((tanx)(tanx)+(1+secx)(1+secx))/(tanx(1+secx))$

$= \frac{{tan}^{2} x + 1 + 2 sec x + {sec}^{2} x}{tan x (1 + sec x)}$ $=(tan^2x+1+2secx+sec^2x)/(tanx(1+secx))$

$= \frac{({tan}^{2} x + 1) + 2 sec x + {sec}^{2} x}{tan x (1 + sec x)}$ $=((tan^2x+1)+2secx+sec^2x)/(tanx(1+secx))$

$= \frac{{sec}^{2} x + 2 sec x + {sec}^{2} x}{tan x (1 + sec x)}$ $=(sec^2x +2secx+sec^2x)/(tanx (1+secx))$

$= \frac{2 {sec}^{2} x + 2 sec x}{tan x (1 + sec x)}$ $=(2sec^2x+2secx)/(tanx(1+secx))$

$= \frac{2 sec x (sec x + 1)}{tan x (1 + sec x)}$ $=(2secx(secx+1))/(tanx(1+secx))$

$= 2 \frac{sec x}{tan x}$ $=2 secx/tanx$

$= 2 \frac{\frac{1}{cos x}}{\frac{sin x}{cos x}}$ $=2 (1/cosx)/(sinx/cosx)$

$= 2 \frac{1}{cos x} \times \frac{cos x}{sin x}$ $=2 1/cosx xx cosx/sinx$

$= 2 \frac{1}{sin x}$ $=2 1/sinx$

$= 2 csc x$ $=2cscx$

$= R H S$ $=RHS$

How do you verify (tan(x)/(1 + sec(x))) + (1+sec(x)/tan(x)) = 2csc(x)(tan(x)1+sec(x))+(1+sec(x)tan(x))=2csc(x)(tan(x)/(1 + sec(x))) + (1+sec(x)/tan(x)) = 2csc(x)?