Question

if y=logtan(π/4+x/2) then show that dy/dx-secx=0
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Answer 1

Answer:

Step by step proof below:

Step-by-step explanation:

Given y = log tan(π/4+x/2)

dy/dx =  $\frac{1}{ tan(\pi /4+x/2)}$ * $\frac{d}{dx}$(tan(π/4+x/2) )                   ($\frac{dy}{dx}$ log x = $\frac{1}{x}$ and Chain Rule)

          = $\frac{1}{ tan(\pi /4+x/2)}$ * sec²(π/4+x/2) * (0+ 1/2)    ($\frac{dy}{dx}$ tan x = sec²x & Chain Rule)

          = $\frac{1}{ tan(\pi /4+x/2)}$ * sec²(π/4+x/2) * (1/2)

          =  $\frac{sec^{2} (\pi /4+x/2) }{2* tan(\pi /4+x/2)}$

          = $\frac{cos (\pi /4+x/2) }{2* sin(\pi /4+x/2)cos^{2} (\pi /4+x/2)}$                  ( tan x = $\frac{sinx}{cos x}$ and sec x = $\frac{1}{cos x}$)

          = $\frac{1 }{2* sin(\pi /4+x/2)cos (\pi /4+x/2)}$

          = $\frac{1 }{ sin 2(\pi /4+x/2)}$                                     ( sin 2x = 2*sin x*cos x)

          = $\frac{1 }{ sin (\pi /2+x)}$                                          

          = $\frac{1 }{ cos x}$                                                   (sin (90 + x) = cos x)

          = sec x                                                 ( sec x = $\frac{1}{cos x}$)

We need to show that $\frac{dy}{dx}$ - sec x = 0

Taking Left Hand Side

=> $\frac{dy}{dx}$ - sec x

Substituting the value of $\frac{dy}{dx}$ from above

= sec x - sec x

= 0

= Right Hand Side

Hence, the result.