Inverse Trigonometric Functions Test - 55

Let $$\displaystyle \tan^{-1} x=\alpha $$
$$\Rightarrow \dfrac{-\pi}{4}<\alpha<0$$
$$\displaystyle \therefore \tan\alpha=x $$ such that $$\displaystyle \frac{-\pi}{4}<\alpha <0 $$
$$\displaystyle \therefore \sin\alpha=\frac{x}{\sqrt{1+x^{2}}}\Rightarrow \alpha=\sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}} \right)$$
$$\displaystyle =\cot^{-1}\left( \frac{1}{x}\right)$$ 
$$\displaystyle=\sec^{-1}\sqrt{1+x^{2}}$$
$$\displaystyle =\ cosec^{-1} \frac{\sqrt{1+x^{2}}}{x}$$ 
$$\displaystyle \alpha=\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$$

(a) $$\rightarrow$$ (4).
4 $$sin^{-1}$$ x + $$\left (\dfrac{\pi}{2} - sin^{-1} x   \right )$$ = x
$$\therefore$$  3$$sin^{-1}$$ x = $$\dfrac{\pi}{2}$$ or x = $$sin\dfrac{\pi}{6} = \dfrac{1}{2}$$
(b) $$\rightarrow$$ (3).
Use Rule 5 and $$a^{2}$$ + $$b^{2}$$ = $$c^{2}$$, $$tan^{-1}$$ 1 = $$\pi/4$$
(c) $$\rightarrow$$ (2).
$$tan^{-1}$$ 2 + $$tan^{-1}$$ 3 = $$\pi$$ + $$tan^{-1}$$ $$\dfrac{2 + 3}{1 - 2.3}$$
= $$\pi$$ - $$tan^{-1}$$ 1
$$\therefore$$ L.H.S = $$tan^{-1}$$ 1 +$$tan^{-1}$$ 2 + $$tan^{-1}$$ 3 = $$\pi$$
(d) $$\rightarrow$$ (1)

$$\sin^{-1}x+\cos^{-1}x$$

$$\sin^{-1}x=\theta_{1}\,\,\Rightarrow \sin\theta_{1}=x$$

$$\cos^{-1}x=\theta_{2}\,\,\Rightarrow \cos\theta_{2}=x$$

$$\tan\theta_{1}=\cfrac{x}{\sqrt{1-x^{2}}}$$

$$\tan\theta_{2}=\cfrac{\sqrt{1-x^{2}}}{x}$$

$$\tan(\theta_{1}+\theta_{2})=\cfrac{\tan\theta_{1}+\tan\theta_{2}}{1-\tan\theta_{1}\tan\theta_{2}}$$

$$=\cfrac{\cfrac{x}{\sqrt{1-x^{2}}}+\cfrac{\sqrt{1-x^{2}}}{x}}{1-1}$$

$$=\cfrac{x^{2}+1-x^{2}}{x\sqrt{1-x^{2}}(0)}$$

$$\tan(\theta_{1}+\theta_{2})=\infty$$

$$\theta_{1}+\theta_{2}=\cfrac{\pi}{2}$$

$$\therefore\,\sin^{-1}x+\cos^{-1}x=\cfrac{\pi}{2}$$

Let,

$$t=2{{\tan }^{-1}}\left[ \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right]$$     ……. (1)

 

Let,

$$b=a\cos \alpha $$      ……. (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

$$ \cos t=\dfrac{\cos \theta +\dfrac{b}{a}}{1+\cos \theta \dfrac{b}{a}}=\dfrac{a\cos \theta +b}{a+b\cos x} $$

$$ \Rightarrow t={{\cos }^{-1}}\left[ \dfrac{a\cos \theta +b}{a+b\cos \theta } \right] $$

 

Hence, this is the required result.