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Question

Prove that:
(i) $$2\tan ^{ -1 }{ \sqrt { \cfrac { b }{ a }  }  } =\cos ^{ -1 }{ \left( \cfrac { a-b }{ a+b }  \right)  } $$
(ii) Find the principal value of $$\cos ^{ -1 }{ \left( -\cfrac { 1 }{ \sqrt { 2 }  }  \right)  } $$


Solution

(i) Let:
$$\theta = \tan^{-1}\sqrt{\dfrac{b}{a}} \Rightarrow \sqrt{\dfrac{b}{a}} = \tan\theta$$
Now, we use the trigonometric identity:
$$\cos 2\theta = \dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$
Substituting for $$\theta$$ and $$\tan\theta$$ and rearranging we get:
$$\cos\left(2\tan^{-1}\sqrt{\dfrac{b}{a}}\right) = \dfrac{a-b}{a+b}$$
$$\Rightarrow 2\tan^{-1}\sqrt{\dfrac{b}{a}} = \cos^{-1}\left(\dfrac{a-b}{a+b}\right)$$
Hence Proved.
(ii) Let $$y$$ be the principal value of $$\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)$$, then $$0\leq y\leq180^o$$
$$\cos(135^o) = -\dfrac{1}{\sqrt{2}} \Rightarrow y = 135^o$$

Mathematics

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