If `(1- tan2^(@) cot 62^(@))/( tan 152^(@) - cot88^(@))=ksqrt(3)` , then value of k is

(1- tan 2 ^(@) cot 62^(@))/(tan 152^(@) -cot 88^(@))=

(1- tan 2 ^(@) cot 62^(@))/(tan 152^(@) -cot 88^(@))=

If (3+cot80^(@)cot 20^(@))/(cot80^(@)+cot20^(@))=tan.(pi)/(k) , then the value of k is (where, (pi)/(k) is an acute angle)

If (1-tan2cot62)/(tan152-cot88)=k sqrt(3) then k=

(1-tan2^(0)cot62^(@))/(tan152^(0)-cot88^(@))(a)sqrt3(b)-sqrt3(c)sqrt2-1(d)1-sqrt2^(@)

If tan a = sqrt(5) - 2 , then the value of tan a - cot a = ?

((1+tan^(2)A)/(1+cot^(2)A))=((1-tan A)/(1-cot A))^(2)=tan^(2)A

If A="tan"^(-1)((xsqrt3)/(2K-x)) and B="tan"^(-1)((2x-K)/(Ksqrt3)) , then the value of A-B is

Without using trigonometric tables , prove that : (i) tan20^(@)tan40^(@) tan45^(@)tan50^(@)tan70^(@)=1 (ii) tan1^(@) tan2^(@)tan60^(@)tan88^(@)tan89^(@)=sqrt(3) (iii) cot5^(@)cot10^(@)cot30^(@)cot80^(@)cot85^(@)=sqrt(3) (iv) 4sin10^(@)sin20^(@)sin30^(@)sec70^(@)sec80^(@)=2