Question

if cos^-1 x +cos^-1 y = π then show that x+y=0​

Answer 1

Solution :

$cos^{-1}x\: +\: cos^{-1}y\:=\: \pi$

$cos^{-1}[xy\:-\:\sqrt{(1\:-\:x^2)(1\: - \: y^2)}] \: =\: \pi$

$xy\:-\:\sqrt{(1\:-\:x^2)(1\: - \: y^2)}\: =\: cos\: \pi$

$xy\:-\:\sqrt{(1\:-\:x^2)(1\: - \: y^2)}\: =\: - 1$

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

Squaring both side

$(xy\: +\: 1)^2 \: = (\: 1 \: - \: y^2 \: -\: x^2\: +\: x^2y^2)^{\frac{1}{\cancel{2}} \times\cancel{2}}$

$\cancel{x^2y^2} \: +\: \cancel{1} \: +\: 2xy \: = \: \cancel{1} \: - \: y^2 \: -\: x^2\: +\: \cancel{x^2y^2}$

$2xy \: +\: x^2 \:+\: y^2\: =\: 0$

$(x\: +\: y)^2\: =\: 0$

$x\: +\: y \: =\: 0$

Hence proved

Final Result

$\boxed{\boxed{\bold{x\:+\:y\: =\: 0}}}$

Answer 2

Answer:

• therefore x+y =0

1. Hope it helps you