Q: What is tan theta minus cot theta?

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cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.

cosine (90- theta) = sine (theta)

Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).

If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)

1 cot(theta)=cos(theta)/sin(theta) cos(45 degrees)=sqrt(2)/2 AND sin(45 degrees)=sqrt(2)/2 cot(45 deg)=cos(45 deg)/sin(deg)=(sqrt(2)/2)/(sqrt(2)/2)=1

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It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.

Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.

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

Yes, it is.

whats the big doubt,cot/tan+1= 1+1= 2

cot theta=tan(90-tetha)

Tan^2

Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).

Since CotΘ = 1 / tanΘ, then tanΘ / cotΘ = tanΘ / (1/tanΘ) = tanΘ x tanΘ = tan²Θ

It is -sqrt(1 + cot^2 theta)

It also equals 13 12.

cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)