Question

show that ,
tan (315°) cot(-405°) + cot (495°) tan (-585°)/sin (-330°) cos(300°) + cos(-390°) sin (420°)​

Answer 1

Answer:

tan (315°) cot(-405°) + cot (495°) tan (-585°)/sin (-330°) cos(300°) + cos(-390°) sin (420°)​ = 2

Step-by-step explanation:

show that ,  

tan (315°) cot(-405°) + cot (495°) tan (-585°)/sin (-330°) cos(300°) + cos(-390°) sin (420°)​ = 2

tan (315°) = tan (360° - 45°) =  -tan45° = -1

cot(-405°) = - Cot(405°) = -Cot(360° + 45°) = -Cot45° = -1

tan (315°) cot(-405°) = (-1)(-1) = 1

cot (495°) = Cot(360° + 135°) = Cot135° = Cot(180 - 45°) = -Cot45° = -1

tan(-585°)= - tan(585°) = -tan(360° + 225°) = - tan(225°) = - tan(180° + 45°)

= - (tan45°) = -tan45° = -1

cot (495°) tan (-585°) = (-1)(-1) = 1

sin (-330°) = -Sin330° = -Sin(360°-30°) = -(-Sin30°) = Sin30°  = 1/2

Cos(300°) = Cos(360° - 60°) = Cos60 = 1/2

sin (-330°) cos(300°)  = (1/2)(1/2) = 1/4

Cos(-390°) = Cos(390°) = Cos(360° + 30°) = Cos30° = √3/2

Sin420° = -Sin(360°+60°) = Sin60° = √3/2

cos(-390°) sin (420°) = (√3/2)(√3/2) = 3/4

putting all these values in LHS

=  (1 + 1)/(1/4 + 3/4)

= 2/1

= 2

= RHS

tan (315°) cot(-405°) + cot (495°) tan (-585°)/sin (-330°) cos(300°) + cos(-390°) sin (420°)​ = 2