Explanation
\tan^{ -1 }( \tan \dfrac { 5\pi }{ 4 } )=\alpha
\Rightarrow \dfrac { 5\pi }{ 4 } =\alpha
Also, \tan^{ -1 }( \tan -\dfrac { 2\pi }{ 3 } )=\beta
\Rightarrow -\dfrac { 2\pi }{ 3 } =\beta
\Rightarrow \alpha +\beta =\dfrac { 5\pi }{ 4 } -\dfrac { 2\pi }{ 3 }
\Rightarrow \alpha +\beta=\dfrac { 7\pi }{ 12 }
\sin^{-1}x+\cos^{-1}x
\sin^{-1}x=\theta_{1}\,\,\Rightarrow \sin\theta_{1}=x
\cos^{-1}x=\theta_{2}\,\,\Rightarrow \cos\theta_{2}=x
\tan\theta_{1}=\cfrac{x}{\sqrt{1-x^{2}}}
\tan\theta_{2}=\cfrac{\sqrt{1-x^{2}}}{x}
\tan(\theta_{1}+\theta_{2})=\cfrac{\tan\theta_{1}+\tan\theta_{2}}{1-\tan\theta_{1}\tan\theta_{2}}
=\cfrac{\cfrac{x}{\sqrt{1-x^{2}}}+\cfrac{\sqrt{1-x^{2}}}{x}}{1-1}
=\cfrac{x^{2}+1-x^{2}}{x\sqrt{1-x^{2}}(0)}
\tan(\theta_{1}+\theta_{2})=\infty
\theta_{1}+\theta_{2}=\cfrac{\pi}{2}
\therefore\,\sin^{-1}x+\cos^{-1}x=\cfrac{\pi}{2}
Let,
t=2{{\tan }^{-1}}\left[ \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right] ……. (1)
b=a\cos \alpha ……. (2)
Then,
\dfrac{a-b}{a+b}=\dfrac{a-a\cos \alpha }{a+a\cos \alpha }
\dfrac{a-b}{a+b}=\dfrac{1-\cos \alpha }{1+\cos \alpha }
\dfrac{a-b}{a+b}=\dfrac{1-\left( 1-2{{\sin }^{2}}\dfrac{\alpha }{2} \right)}{1+2{{\cos }^{2}}\dfrac{\alpha }{2}}
\dfrac{a-b}{a+b}=\dfrac{2{{\sin }^{2}}\dfrac{\alpha }{2}}{2{{\cos }^{2}}\dfrac{\alpha }{2}}
\dfrac{a-b}{a+b}={{\tan }^{2}}\dfrac{\alpha }{2}
\sqrt{\dfrac{a-b}{a+b}}=\tan \dfrac{\alpha }{2} …… (3)
From equation (1) and (3), we get
\tan \dfrac{t}{2}=\tan \dfrac{\alpha }{2}\tan \dfrac{\theta }{2}
\Rightarrow \cos t=\dfrac{1-{{\tan }^{2}}\dfrac{t}{2}}{1+{{\tan }^{2}}\dfrac{t}{2}}
\cos t=\dfrac{1-{{\tan }^{2}}\dfrac{\alpha }{2}{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\alpha }{2}{{\tan }^{2}}\dfrac{\theta }{2}}
\cos t=\dfrac{1-\dfrac{{{\sin }^{2}}\dfrac{\alpha }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}}\dfrac{{{\sin }^{2}}\dfrac{\theta }{2}}{{{\cos }^{2}}\dfrac{\theta }{2}}}{1+\dfrac{{{\sin }^{2}}\dfrac{\alpha }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}}\dfrac{{{\sin }^{2}}\dfrac{\theta }{2}}{{{\cos }^{2}}\dfrac{\theta }{2}}}
\cos t=\dfrac{{{\cos }^{2}}\dfrac{\alpha }{2}{{\cos }^{2}}\dfrac{\theta }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}{{\sin }^{2}}\dfrac{\theta }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}{{\cos }^{2}}\dfrac{\theta }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}{{\sin }^{2}}\dfrac{\theta }{2}}
\cos t=\dfrac{\dfrac{1}{2}\left[ \cos \theta +\cos \alpha \right]}{\dfrac{1}{2}\left[ 1+\cos \alpha \cos \theta \right]}
From equation (2), we get
\cos t=\dfrac{\cos \theta +\dfrac{b}{a}}{1+\cos \theta \dfrac{b}{a}}=\dfrac{a\cos \theta +b}{a+b\cos x}
\Rightarrow t={{\cos }^{-1}}\left[ \dfrac{a\cos \theta +b}{a+b\cos \theta } \right]
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