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How do you simplify tanx plus tany per cosx plus coty?
(tan x- 1)/ (1+tan x)
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
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
What is cot theta divided by tan theta plus one equal?
whats the big doubt,cot/tan+1= 1+1= 2
Tan 9 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
Cos2x equals 1 minus tan squared x divide by1 plus tan squared x?
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
How can you verify 1 plus tan theta divided by 1 minus tan theta equals cot theta plus 1 divided by cot theta minus 1?
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
How do you simplify tanx plus tany per cosx plus coty?
(tan x- 1)/ (1+tan x)
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
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
What is cot theta divided by tan theta plus one equal?
whats the big doubt,cot/tan+1= 1+1= 2
If cosA plus cosB 12 and sinA plus sinB 14 then tan(A2 plus B2)?
1 by 2
If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
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
What is tan20tan32 plus tan32tan38 plus tan38tan20?
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
Tan 9 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
Cos2x equals 1 minus tan squared x divide by1 plus tan squared x?
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
What is tan20 tan31 plus tan32 tan38 plus tan 38 tan 20?
It is approx 0.9725
Tan 9-tan 27-tan 63 plus tan81 equals 4?
4
What is the exact trigonometric function value of cot 15 degrees?
cot(15)=1/tan(15) Let us find tan(15) tan(15)=tan(45-30) tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b)) tan(45-30)= (tan(45)-tan(30))/(1+tan(45)tan(30)) substitute tan(45)=1 and tan(30)=1/√3 into the equation. tan(45-30) = (1- 1/√3) / (1+1/√3) =(√3-1)/(√3+1) The exact value of cot(15) is the reciprocal of the above which is: (√3+1) /(√3-1)
