28. [ 17 ] Let w , x , y , and z be positive real numbers such that 0 6 = cos w cos x cos y cos z 2 π = w + x + y + z 3 tan w = k (1 + sec w ) 4 tan x = k (1 + sec x ) 5 tan y = k (1 + sec y ) 6 tan z = k (1 + sec z ) . 1 (Here sec t denotes when cos t 6 = 0.) Find k . cos t √ u sin u w x Answer: 19 From the identity tan = , the conditions work out to 3 tan = 4 tan = 2 1+cos u 2 2 y y z w x z 5 tan = 6 tan = k . Let a = tan , b = tan , c = tan , and d = tan . Using the identity 2 2 2 2 2 2 tan M +tan N tan( M + N ) = , we obtain 1 − tan M tan N ( ) ( ) ( ) y + z w + x tan + tan w + x y + z 2 2 ( ) ( ) tan + = y + z w + x 2 2 1 − tan tan 2 2 a + b c + d

1 − ab 1 − cd ( ) ( )

a + b c + d 1 − 1 − ab 1 − cd a + b + c + d − abc − abd − bcd − acd = . 1 + abcd − ab − ac − ad − bc − bd − cd ( ) x + y + z + w Because x + y + z + w = π , we get that tan = 0 and thus a + b + c + d = abc + abd + bcd + acd . 2 3 Substituting a, b, c, d corresponding to the variable k , we obtain that k − 19 k = 0. Therefore, k can √ √ √ be only 0 , 19 , − 19. However, k = 0 is impossible as w, x, y, z will all be 0. Also, k = − 19 is √ impossible as w, x, y, z will exceed π . Therefore, k = 19.