# Thread: Inverse trigonometric Function Question 2

1. ## Inverse trigonometric Function Question 2

Prove that
arc sinx + arc cosx = arc tan [xy+{(1-x^2)(1-y^2)}^1/2]/[y{(1-x^2)}^1/2 -x{(1-y^2)}^1/2]

2. ## Trig equation

Hello tariq_h_tauheed
Originally Posted by tariq_h_tauheed
Prove that
arc sinx + arc cosx = arc tan [xy+{(1-x^2)(1-y^2)}^1/2]/[y{(1-x^2)}^1/2 -x{(1-y^2)}^1/2]
I'm sure you mean $\arcsin x + \arccos y= ...$.

So let $a = \arcsin x, b = \arccos y$.

$\Rightarrow x = \sin a,\quad y = \cos b$

$\Rightarrow \cos a = (1-\sin^2a)^{\tfrac12} = (1-x^2)^{\tfrac12}, \quad\sin b = (1-y^2)^{\tfrac12}$

Then $\tan(a+b) = \frac{\sin(a+b)}{\cos(a+b)}$

$=\frac{\sin a\cos b+\cos a\sin b}{\cos a\cos b - \sin a \sin b}$

$\Rightarrow \arcsin x + \arccos y = a + b = \arctan\Big(\frac{\sin a\cos b+\cos a\sin b}{\cos a\cos b - \sin a \sin b}\Big)$

$=\arctan\Bigg(\frac{xy+(1-x^2)^{\tfrac12}(1-y^2)^{\tfrac12}}{y(1-x^2)^{\tfrac12}-x(1-y^2)^{\tfrac12}}\Bigg)$

3. Hello, tariq_h_tauheed!

Another approach . . .

Prove that: . $\arcsin x + \arccos y \:=\: \arctan \left[\frac{xy+\sqrt{(1-x^2)(1-y^2)}} {y\sqrt{1-x^2} -x\sqrt{1-y^2}}\right]$

Let: . $\begin{array}{ccccccc}\alpha \:=\:\arcsin x & \Rightarrow & \sin\alpha \:=\:x \\
\beta \:=\:\arccos y & \Rightarrow & \cos\beta \:=\:y \end{array}$

Then we have: . $\tan\alpha \;=\;\frac{x}{\sqrt{1-x^2}} \qquad \tan\beta \;=\; \frac{\sqrt{1-y^2}}{y}$ .[1]

The left side is: . $\alpha + \beta$

Take the tangent: . $\tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$ .[2]

Substitute [1] into [2]: . $\tan(\alpha + \beta) \;=\;\frac{\dfrac{x}{\sqrt{1-x^2}} + \dfrac{\sqrt{1-y^2}}{y}} {1 - \dfrac{x}{\sqrt{1-x^2}}\cdot\dfrac{\sqrt{1-y^2}}{y}}$

Multiply top and bottom by $\left(y\sqrt{1-x^2}\right)$

. . $\tan(\alpha + \beta) \;=\;\frac{y\sqrt{1-x^2}\left(\frac{x}{\sqrt{1-x^2}} + \frac{\sqrt{1-y^2}}{y}\right)} {y\sqrt{1-x^2}\left(1 - \frac{x}{\sqrt{1-x^2}}\cdot\frac{\sqrt{1-y^2}}{y}\right)}\;=$ . $\frac{xy + \sqrt{1-x^2}\sqrt{1-y^2}}{y\sqrt{1-x^2} - x\sqrt{1-y^2}}$

Therefore: . $\alpha + \beta \;=\;\arcsin x + \arccos y \;=\;\arctan\left[\frac{xy + \sqrt{1-x^2}\sqrt{1-y^2}}{y\sqrt{1-x^2} - x\sqrt{1-y^2}}\right]$

4. Thanks a ton...