2sqrt(2)sin10^(@)[(sec5^(@))/(2)+(cos40^(@))/(sin5^(@))-2sin35']

Calculate the following without using trigonometric tables (i) tan 9^(@) - tan 27^(@) - tan 63^(@) + tan 81^(@) (ii) cosec 10^(@) -sqrt(3) sec 10^(@) (iii) 2sqrt(2) sin 10^(@) [(sec 5^(@))/(2) + (cos 40^(@))/(sin 5^(@))-2 sin 35^(@)] (iv) cot 70^(@)+4 cos 70^(@)

Find the value 2 sqrt 2 sin 10^0[(s e c5^0)/2+(cos40^0\ )/(sin5^0)-2sin 35^0]=

cos20^(@)cos40^(@)-sin5^(@)sin25^(@)=

Find the least positive x in degree for which tan x= (cos5^(@)cos20^(@)+cos35^(@)cos50^(@)-sin5^(@)sin20^(@)-sin35^(@)sin50^(@))/(sin5^(@)cos20^(@)-sin35^(@)cos50^(@)+cos5^(@)sin20^(@)-cos35^(@)sin50^(@))

The smallest positive value of x (in degrees) for whichcos tan x=(cos5^(@)cos20^(@)+cos35^(@)cos50^(@)-sin50^(@)sin20^(@)-sin35^(@)sin50^(@))/(sin5^(@)cos20^(@)-sin35^(@)cos50^(@)+cos5^(@)sin20^(@)-cos35^(@)sin50^(@))

((sin 35^(@))/(cos 55^(@)))^(2)-((cos55^(@))/(sin 35^(@)))^(2) +2 sin 30^(@) is equal to

2 sin 40^(@) sin 10^(@)=

sin20^(@)*cos40^(@)+cos20^(@)*sin40^(@)=(sqrt(3))/(2)

(sin2^(@)+sin4^(@)+sin6^(@)+sin8^(@))/(cos1^(@)*cos2^(@)*cos5^(@))=