Question

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

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

${tan}^{-1}\sqrt{\left\{\frac{x(x+y+z)}{yz}\right\}}+{tan}^{-1}\sqrt{\left\{\frac{y(x+y+z)}{zx}\right\}}+{tan}^{-1}\sqrt{\left\{\frac{z(x+y+z)}{xy}\right\}}=$

• A. $\pi$
• B. $4\pi$
• C. $3\pi$
• D. $2\pi$

Answer 1

The correct option is A $\pi$

(a) Put $x+y+z=r$

$\therefore$ ${tan}^{-1}\sqrt{\left(\frac{rx}{yz}\right)}+{tan}^{-1}\sqrt{\left(\frac{ry}{zx}\right)}$

$={tan}^{-1}\frac{\sqrt{r/xyz}.(x+y)}{(1-r/z)}$

$={tan}^{-1}\frac{\sqrt{\left(rz\right)/\left(xy\right)}.(x+y)}{-(x+y)}$

$=\pi +{tan}^{-1}\{-\sqrt{\left(\frac{rz}{xy}\right)}\}$

$=\pi -{tan}^{-1}\sqrt{\left(\frac{rz}{xy}\right)}$

$\therefore$ ${tan}^{-1}\left(\frac{rz}{yz}\right)+{tan}^{-1}\sqrt{\left(\frac{ry}{zx}\right)}+{tan}^{-1}\sqrt{\left(\frac{rz}{xy}\right)}=\pi$

(b) We have to prove that

${tan}^{-1}\frac{yz}{xr}+{tan}^{-1}\frac{zr}{yr}+{tan}^{-1}\frac{xy}{zr}=\frac{\pi }{2}$

or ${tan}^{-1}\frac{\left(z/r\right)\left[\right(x^2+y^2\left)/\right(xy\left)\right]}{1-\left(z^2/r^2\right)}=\frac{\pi }{2}-{tan}^{-1}\frac{xy}{zr}$

or ${tan}^{-1}\frac{\left(zr/xy\right)(x^2+y^2)}{x^2+y^2}={cot}^{-1}\frac{xy}{zr}$

or ${ran}^{-1}\frac{zr}{xy}={tan}^{-1}\frac{zr}{xy}$, which is true.