It is a meaningless expression. Tan is a function which requires an angle as its argument.
(tan x- 1)/ (1+tan x)
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
whats the big doubt,cot/tan+1= 1+1= 2
tan(9) + tan(81) - tan(27) - tan(63) = 4
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
(tan x- 1)/ (1+tan x)
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
whats the big doubt,cot/tan+1= 1+1= 2
1 by 2
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
tan(9) + tan(81) - tan(27) - tan(63) = 4
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
It is approx 0.9725
This may not be the most efficient method but ... Let the three angle be A, B and C. Then note that A + B + C = 20+32+38 = 90 so that C = 90-A+B. Therefore, sin(C) = sin[(90-(A+B) = cos(A+B) and cos(C) = cos[(90-(A+B) = sin(A+B). So that tan(C) = sin(C)/cos(C) = cos(A+B) / sin(A+B) = cot(A+B) Now, tan(A+B) = [tan(A)+tan(B)] / [1- tan(A)*tan(B)] so cot(A+B) = [1- tan(A)*tan(B)] / [tan(A)+tan(B)] The given expressin is tan(A)*tan(B) + tan(B)*tan(C) + tan(C)*tan(A) = tan(A)*tan(B) + [tan(B) + tan(A)]*cot(A+B) substituting for cot(A+B) gives = tan(A)*tan(B) + [tan(B) + tan(A)]*[1- tan(A)*tan(B)]/[tan(A)+tan(B)] cancelling [tan(B) + tan(A)] and [tan(A) + tan(B)], which are equal, in the second expression. = tan(A)*tan(B) + [1- tan(A)*tan(B)] = 1
4
Tan(3a_)(sqrt(2) / 2 )(sqrt(2)/2) + 0.5 Tan(3a) ( 2/4) + 1/2 Tan(3a) ( 1/2 + 1/2 ) Sin(3a) (1) / 2Cos(3a) + 1/2 Without beinf equated to anything this will not go any further.