Question

u= log tan(π/4+x/2) prove that, sinh u = tan x​

Answer 1

Answer:

Given, u=logtan(

4

π

+

2

θ

)

tanh(

2

u

)=

cosh(

2

u

)

sinh(

2

u

)

=

e

2

u

+e

−

2

u

e

2

u

−e

−

2

u

=

tan(

4

π

)+

cot(

4

π

+

2

θ

)

tan(

4

π

+

2

θ

)

−

cot(

4

π

+

2

θ

)

Answer 2

Answer:

Proof is below:

Step-by-step explanation:

Given that u = log tan($\frac{\pi}{4}$ + $\frac{x}{2}$)

Taking e both sides, we get

$e^{u}$ =  tan($\frac{\pi}{4}$ + $\frac{x}{2}$)

$e^{u}$ = $\frac{tan\frac{\pi}{4} + tan\frac{x}{2}}{1 -tan\frac{\pi}{4} tan\frac{x}{2} }$

=> $e^{u}$ = $\frac{1 + tan\frac{x}{2}}{1 -1* tan\frac{x}{2} }$

We need to prove sinh u = tan x

We know that sinh u =  $\frac{e^{u}-e^{-u}}{2}$ =  $\frac{e^{2u}-1}{2e^{u}}$  

2 $e^{u}$ = $2\left(\frac{1 + tan\frac{x}{2}}{1 -tan\frac{x}{2} }\right)$

Calculating  $\frac{e^{2u}-1}{2e^{u}}$

Substituting the values, we get,

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

Taking LCM

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

Cancelling the common terms,

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

Simplifying using (a ± b)² = a² + b² ± 2ab

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

Cancelling the common terms

=  $\frac{2 tan \frac{x}{2}}{1-tan^2\frac{x}{2}}$

By double angle formula of tan x

= tan (2*$\frac{x}{2}$)

= tan x

Hence proved.