Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.

${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$

$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.

(a) If ${tan}^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\surd (1+x^2)+\surd (1+x^2)}=\alpha$. then prove that $x^2=sin2\alpha$

(b) If $\frac{mtan(\alpha -\theta )}{{cos}^2\theta }=\frac{ntan\theta }{{cos}^2(\alpha -\theta )}$, then prove that $\theta =\frac{1}{2}[\alpha -{tan}^{-1}\left(\frac{n-m}{n+m}tan\alpha \right)]$

(c) ${cos}^{-1}\frac{cos\alpha +cos\beta }{1+cos\alpha cos\beta }=2{tan}^{-1}\left(tan\frac{\alpha }{2}tan\frac{\beta }{2}\right)$

Answer 1

(a) From the given question

$\frac{\sqrt{1+x^2}-\sqrt{1+x^2}}{\surd (1+x^2)+\surd (1-x^2)}=tan\alpha =\frac{sin\alpha }{cos\alpha }$.

Apply componendo and dividendo

$\frac{2\sqrt{1+x^2}}{2\surd (1-x^2)}=\frac{cos\alpha +sin\alpha }{cos\alpha -sin\alpha }Square$

$\frac{1+x^2}{1-x^2}=\frac{1+sin2\alpha }{1-sin2\alpha }$, $\because$ ${cos}^2\alpha +{sin}^2\alpha =1$

and $2sin\alpha cos\alpha =sin2\alpha$.

Hence $x^2=sin2\alpha$ from above.

Alternative : Put $x^2=cos\theta$. Rest do yourself or put $x^2=sin2\alpha$

$\therefore$ $\sqrt{1+x^2}=\sqrt{1+sin2\alpha }=cos\alpha +sin\alpha$ etc.

(b) From the given relation, we get

$\frac{m}{n}=\frac{{cos}^2\theta tan\theta }{tan(\alpha -\theta ){cos}^2(\alpha -\theta )}$

$=\frac{2sin\theta cos\theta }{2sin(\alpha -\theta )cos(\alpha -\theta )}=\frac{sin2\theta }{sin(2\alpha -2\theta )}$

Now applying componendo and dividendo, we get

$\frac{n+m}{n-m}=\frac{sin(2\alpha -2\theta )+sin2\theta }{sin(2\alpha -2\theta )-sin2\theta }=\frac{2sin\alpha cos(\alpha -2\theta )}{2cos\alpha sin(\alpha -2\theta )}$

or $tan(\alpha -2\theta )=\frac{n-m}{n+m}tan\alpha$

$\therefore$ $\alpha -2\theta ={tan}^{-1}\left(\frac{n-m}{n+m}tan\alpha \right)$

or $\theta =\frac{1}{2}[\alpha -{tan}^{-1}\left(\frac{n-m}{n+m}tan\alpha \right)]$.

(c) Let ${cos}^{-1}\frac{cos\alpha +cos\beta }{1+cos\alpha cos\beta }=\theta$

$\therefore$ $\frac{cos\theta }{1}=\frac{cos\alpha +cos\beta }{1+cos\alpha cos\beta }$

Apply componendo and dividendo

$\frac{1-cos\theta }{1+cos\theta }=\frac{(1-cos\alpha )(1-cos\beta )}{(1+cos\alpha )(1+cos\beta )}$

or ${tan}^2\frac{\theta }{2}={tan}^2\frac{\alpha }{2}{tan}^2\frac{\beta }{2}$

$\therefore$ $tan\frac{\theta }{2}=tan\frac{\alpha }{2}tan\frac{\beta }{2}$

$\therefore$ $\theta =2{tan}^{-1}\left(tan\frac{\alpha }{2}tan\frac{\beta }{2}\right)$