More fun questions :) Trigonometry and Complex Numbers

1. Prove that 

$\tan{(~\arccos{(x)}~)}=\cot{(~\arcsin{(x)}~)}$

i)

$\begin{array}{lrcll} & \cos{(\varphi)} &=& x \\ \text{or}& \quad \varphi &=& \arccos{(x)}\\ \end{array}$

ii)

$\begin{array}{lrcll} &\sin{(90^\circ-\varphi)} &=& \cos{(\varphi)} = x \\ \text{or}& \quad 90^\circ-\varphi &=& \arcsin{(x)}\\ \end{array}$

iii)

$\begin{array}{rcll} \cot{(90^\circ-\varphi)} &=& \tan{(\varphi)} \\ \cot{(\arcsin{(x)})} &=& \tan{(\arccos{(x)})} \\ \end{array}$

2)

Solve for x over the real numbers:
log(x + 1) = log(2) + (log(x))/2

Subtract log(2) + (log(x))/2 from both sides:
-log(2) - (log(x))/2 + log(x + 1) = 0

Bring -log(2) - (log(x))/2 + log(x + 1) together using the common denominator 2:
1/2 (-2 log(2) - log(x) + 2 log(x + 1)) = 0

Multiply both sides by 2:
-2 log(2) - log(x) + 2 log(x + 1) = 0

-2 log(2) - log(x) + 2 log(x + 1) = log(1/4) + log(1/x) + log((x + 1)^2) = log((x + 1)^2/(4 x)):
log((x + 1)^2/(4 x)) = 0

Cancel logarithms by taking exp of both sides:
(x + 1)^2/(4 x) = 1

Multiply both sides by 4 x:
(x + 1)^2 = 4 x

Subtract 4 x from both sides:
(x + 1)^2 - 4 x = 0

Expand out terms of the left hand side:
x^2 - 2 x + 1 = 0

Write the left hand side as a square:
(x - 1)^2 = 0

Take the square root of both sides:
x - 1 = 0

Add 1 to both sides:
Answer: |x = 1

1)

Verify the following identity:
tan(cos^(-1)(θ)) = cot(sin^(-1)(θ))

Multiply numerator and denominator of sqrt(1 - θ^2)/θ by 1/sqrt(1 - θ^2):
1/(θ 1/sqrt(1 - θ^2)) = ^?(sqrt(1 - θ^2))/(θ)

1 = 1:
1/(θ 1/sqrt(1 - θ^2)) = ^?(sqrt(1 - θ^2))/(θ)

θ/sqrt(1 - θ^2) = θ/sqrt(1 - θ^2):
1/θ 1/sqrt(1 - θ^2) = ^?(sqrt(1 - θ^2))/(θ)

The left hand side and right hand side are identical:
Answer: |(identity has been verified)

I have to sleep, I will give the answer for 2..

You probably know that: $\ln z =\ln|z| +i \arg(z)$

If you try $\ln i$:

$\ln i\\ = \ln |i| + i \arg(i)\\ = \ln 1 + i \arctan \dfrac{1}{0}\\ =\ln 1 + i \dfrac{\pi}{2}\\ = \dfrac{i\pi}{2}$

If you try $\ln(1+i)$:

$\ln(1+i)\\ =\ln|1+i| + i \arg(1+i)\\ =\ln 2 + i \arctan{1}\\ =\ln 2 + \dfrac{i\pi}{4}\\ = \ln 2 + \dfrac{\ln i}{2}$

Therefore x = i satisfies the equation.