Question

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

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

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

If $u={cot}^{-1}\left[\sqrt{cos2\theta }\right]-{tan}^{-1}\left[\sqrt{cos2\theta }\right]$, then prove that $sinu={tan}^2\theta$.

Answer 1

we have $u={cot}^{-1}\sqrt{cos2\theta }-{tan}^{-1}\sqrt{cos2\theta }$

$={tan}^{-1}\left(1/\sqrt{cos2\theta }\right)-{tan}^{-1}\sqrt{cos2\theta }$

$={tan}^{-1}\frac{1/\left(\sqrt{cos2\theta }\right)-\sqrt{cos2\theta }}{1+[1/\surd 9cos2\theta )]\surd (cos2\theta )}$

$={tan}^{-1}\frac{1-cos2\theta }{2\surd \left(cos2\theta \right)}$

$\therefore$ $tanu=\frac{1-cos2\theta }{2\surd \left(cos2\theta \right)}$

or $cotu=\frac{2\sqrt{cos2\theta }}{1-cos2\theta }$

Hence ${cosec}^{2}u=1+{cot}^{2}u$

$=1+\frac{4cos2\theta }{{(1-cos2\theta )}^2}=\frac{{(1+cos2\theta )}^2}{{(1-cos2\theta )}^2}$

or $cosecu=\frac{1+cos2\theta }{1-cos2}=\frac{2{cos}^2\theta }{2{sin}^2\theta }={cot}^2\theta$

or $sinu={tan}^2\theta$.