To prove:- tan{pi/4 + cos^-1(a/b)} + tan{pi/4 - cos^-1(a/b)} =2b/a
P poras21 New member Joined Jan 26, 2012 Messages 1 Jan 26, 2012 #1 To prove:- tan{pi/4 + cos^-1(a/b)} + tan{pi/4 - cos^-1(a/b)} =2b/a Last edited: Jan 26, 2012
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jan 26, 2012 #2 Hello, poras21! Please check the problem . . . it is not an identity. \(\displaystyle \text{Prove: }\:\tan\left[\frac{\pi}{4} + \cos^{\text{-}1}(\frac{a}{b})\right] + \tan\left[\frac{\pi}{4} - \cos^{\text{-}1}(\frac{a}{b})\right] \:=\:\frac{2b}{a}\) Click to expand... \(\displaystyle \text{Let }\theta \,=\,\cos^{-1}(\frac{a}{b}) \quad\Rightarrow\quad \cos\theta \,=\,\frac{a}{b} \,=\,\frac{adj}{hyp} \) . . \(\displaystyle \text{Then: }\,opp \,=\,\sqrt{b^2-a^2} \quad\Rightarrow\quad \tan\theta \,=\,\frac{opp}{adj} \,=\,\frac{\sqrt{b^2-a^2}}{a}\) \(\displaystyle \tan\left[\frac{\pi}{4} + \cos^{\text{-}1}(\frac{a}{b})\right] + \tan\left[\frac{\pi}{4} - \cos^{\text{-}1}(\frac{a}{b})\right] \) . . \(\displaystyle =\;\dfrac{\tan\frac{\pi}{4} + \tan\left[\cos^{-1}(\frac{a}{b})\right]}{1 - \tan\frac{\pi}{4}\tan\left[\cos^{-1}(\frac{a}{b})\right]} + \dfrac{\tan\frac{\pi}{4} - \tan\left[\cos^{-1}(\frac{a}{b})\right]}{1 + \tan\frac{\pi}{4}\tan\left[\cos^{-1}(\frac{a}{b})\right]}\) . . \(\displaystyle =\;\dfrac{1 + \dfrac{\sqrt{b^2-a^2}}{a}}{1 - \dfrac{\sqrt{b^2-a^2}}{a}} + \dfrac{1 - \dfrac{\sqrt{b^2-a^2}}{a}}{1 + \dfrac{\sqrt{b^2-a^2}}{a}}\) . . \(\displaystyle =\;\dfrac{a + \sqrt{b^2-a^2}}{a - \sqrt{b^2-a^2}} + \dfrac{a - \sqrt{b^2-a^2}}{a + \sqrt{b^2-a^2}}\) . . \(\displaystyle =\;\dfrac{(a+\sqrt{b^2-a^2})^2 + (a-\sqrt{b^2-a^2})^2}{a^2 - (b^2-a^2)} \) . . \(\displaystyle =\;\dfrac{\left[a^2 + 2a\sqrt{b^2-a^2} + (b^2-a^2)\right] + \left[a^2 - 2a\sqrt{b^2-a^2} + (b^2-a^2)\right]}{2a^2-b^2}\) . . \(\displaystyle =\;\dfrac{2b^2}{2a^2-b^2}\)
Hello, poras21! Please check the problem . . . it is not an identity. \(\displaystyle \text{Prove: }\:\tan\left[\frac{\pi}{4} + \cos^{\text{-}1}(\frac{a}{b})\right] + \tan\left[\frac{\pi}{4} - \cos^{\text{-}1}(\frac{a}{b})\right] \:=\:\frac{2b}{a}\) Click to expand... \(\displaystyle \text{Let }\theta \,=\,\cos^{-1}(\frac{a}{b}) \quad\Rightarrow\quad \cos\theta \,=\,\frac{a}{b} \,=\,\frac{adj}{hyp} \) . . \(\displaystyle \text{Then: }\,opp \,=\,\sqrt{b^2-a^2} \quad\Rightarrow\quad \tan\theta \,=\,\frac{opp}{adj} \,=\,\frac{\sqrt{b^2-a^2}}{a}\) \(\displaystyle \tan\left[\frac{\pi}{4} + \cos^{\text{-}1}(\frac{a}{b})\right] + \tan\left[\frac{\pi}{4} - \cos^{\text{-}1}(\frac{a}{b})\right] \) . . \(\displaystyle =\;\dfrac{\tan\frac{\pi}{4} + \tan\left[\cos^{-1}(\frac{a}{b})\right]}{1 - \tan\frac{\pi}{4}\tan\left[\cos^{-1}(\frac{a}{b})\right]} + \dfrac{\tan\frac{\pi}{4} - \tan\left[\cos^{-1}(\frac{a}{b})\right]}{1 + \tan\frac{\pi}{4}\tan\left[\cos^{-1}(\frac{a}{b})\right]}\) . . \(\displaystyle =\;\dfrac{1 + \dfrac{\sqrt{b^2-a^2}}{a}}{1 - \dfrac{\sqrt{b^2-a^2}}{a}} + \dfrac{1 - \dfrac{\sqrt{b^2-a^2}}{a}}{1 + \dfrac{\sqrt{b^2-a^2}}{a}}\) . . \(\displaystyle =\;\dfrac{a + \sqrt{b^2-a^2}}{a - \sqrt{b^2-a^2}} + \dfrac{a - \sqrt{b^2-a^2}}{a + \sqrt{b^2-a^2}}\) . . \(\displaystyle =\;\dfrac{(a+\sqrt{b^2-a^2})^2 + (a-\sqrt{b^2-a^2})^2}{a^2 - (b^2-a^2)} \) . . \(\displaystyle =\;\dfrac{\left[a^2 + 2a\sqrt{b^2-a^2} + (b^2-a^2)\right] + \left[a^2 - 2a\sqrt{b^2-a^2} + (b^2-a^2)\right]}{2a^2-b^2}\) . . \(\displaystyle =\;\dfrac{2b^2}{2a^2-b^2}\)
