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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More