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If x,y,z are are acute angles and cosx=tany,cosy=tanz,cosz=tanx then 1234sinx+1342siny+1423sinz=ksin18 . where k is_

If x,y,z are are acute angles and cosx=tany,cosy=tanz,cosz=tanx then 1234sinx+1342siny+1423sinz=ksin18.where k is_

Grade:Upto college level

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
576-2133_1.PNG576-2234_2.PNG576-2021_3.PNG576-1357_4.PNG
Vikas TU
14149 Points
one year ago
Dear student 
We have : cos x = tan y. 
∴ cos² x = tan² y = sec² y - 1 ............... (1) 
But cos y = tan z ∴ sec y = cot z 
∴ from (1), 
... cos² x = cot² z - 1 
∴ 1 + cos² x = cot² z = cos² z / sin² z = cos² z / ( 1 - cos² z ) 
But cos z = tan x 
∴ 1 + cos² x = tan² x / ( 1 - tan² x ) 
∴ 1 + ( 1 - sin² x ) = ( sin² x / cos² x ) / [ 1 - ( sin² x / cos² x ) ] 
∴ 2 - sin² x = sin² x / ( cos² x - sin² x ) 
∴ 2 - sin² x = sin² x / ( 1 - 2 sin² x ) 
∴ ( 2 - sin² x )( 1 - 2 sin² x ) = sin² x 
∴ 2 sin⁴ x - 6 sin² x + 2 = 0 
∴ by Quadratic Formula, 
sin² x = [ 3 ± √(9-4) ] / 2 = ( 3 ± √5 ) / 2 
But ( 3 + √5 ) /2 > 1 whereas sin² x ≤ 1. 
∴ sin² x = ( 3 - √5 ) / 2 = ( 6 - 2√5 ) / 4 = ( √5 - 1 )² / 4 
∴ sin x = ( √5 - 1 ) / 2 = 2 sin 18° 
We can similarly show that 
... sin x = sin y = sin z = 2 sin 18° = ( √5 - 1 ) / 2

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