The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
It also equals 13 12.
A = Root (Q squared plus P squared) C = 90 + tan inverse P/Q ... I think lol just worked it out just now =w=
Use tangent. Your equation will be tan(slope of hypotenuse) = opposite side / adjacent side. it's easier if you just do A squared plus b squared equals c squared. Then subtitute the numbers gived in.
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
Yes.
22
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
It also equals 13 12.
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
A = Root (Q squared plus P squared) C = 90 + tan inverse P/Q ... I think lol just worked it out just now =w=
4
Tan
Use tangent. Your equation will be tan(slope of hypotenuse) = opposite side / adjacent side. it's easier if you just do A squared plus b squared equals c squared. Then subtitute the numbers gived in.
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
Tan^2
Yes, it is.