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 ${cos}^{-1}p+{cos}^{-1}q+{cos}^{-1}r=\pi$, then prove that

$p^2+q^2+r^2+2pqr=1$

(b) If ${sin}^{-1}x+{sin}^{-1}y+{sin}^{-1}z=\pi$, then prove that

$x^4+y^4+z^4+4x^2{y}^2{z}^2=2(x^2{y}^2+y^2{z}^2+z^2{x}^2)$

(c) If ${tan}^{-1}x+{tan}^{-1}y+{tan}^{-1}z=\pi$ or $\pi /2$ show that

$x+y+z=xyz$or $xy+yz+zx=1$.

Answer 1

(a) ${cos}^{-1}p+{cos}^{-1}q+{cos}^{-1}r=\pi$

$\Rightarrow$ ${cos}^{-1}[pq-\sqrt{1-p^2}.\sqrt{1-q^2}]=\pi -{cos}^{-1}r={cos}^{-1}(-r)$.

$\therefore$ $pq-\sqrt{(1-p^2)(1-q^2)}=-r$

or ${(pq+r)}^2=(1-p^2)(1-q^2)$

or $p^2{q}^2+r^2+2pqr=1-p^2-q^2+p^2{q}^2$

or $p^2+q^2+r^2+2pqr=1$.

(b) ${sin}^{-1}x+{sin}^{-1}y=\pi -{sin}^{-1}z$

Take cos of both sides and not sine for easy simplification.

$cos\left({sin}^{-1}t\right)=cos\left({cos}^{-1}\sqrt{1-t^2}\right)=\sqrt{(1-t^2)}$

$\sqrt{1-x^2}\sqrt{1-y^2}-xy=-\sqrt{1-z^2}$

$\therefore$ $(1-x^2(1-y^2)=x^2{y}^2+(1-z^2)-2xy\sqrt{1-z^2}$

$1-x^2-y^2=1-z^2-2xy\sqrt{1-z^2}$

$\therefore$ $x^2+y^2-z^2=2xy\sqrt{1-z^2}$, Square again

(c) L.H.S. $={tan}^{-1}\frac{(x+y)}{1-xy}+{tan}^{-1}z$

$={tan}^{-1}\frac{\left[\right(x+y\left)/\right(1-xy\left)\right]+z}{1-\left[\right(x+y\left)/\right(1-xy\left)\right],z}$

$={tan}^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}=\pi$ or $\frac{\pi }{2}$

$\therefore$ $\frac{x+y+z-xyz}{1-xy-yz-zx}=tan\pi$ or $tan\frac{\pi }{2}$

or $=0$ or $\infty$

Hence $x+y+z-xyz=0$ i.e. $N^r=0$

i.e. $x+y+z=xyz$

or $1-xy-yz-zx=0$ i.e. $D^r=0$

i.e. $xy+yz+zx=1$