cotx+tan2x=cotxsec2x
prove
cosx/sinx + 2sinxcosx/cos2x
= (cosxcos2x + 2sin^2(x) cosx)/(sinxcos2x)
= cotxsec2x (cos2x + 2sin^2(x))
= cotx sec2x (1-2sin^2(x) + 2sin^2(x))
= cotx sec2x
To prove that cot(x) + tan(2x) = cot(x)sec^2(x), we need to manipulate the given expression using trigonometric identities.
We'll start with the left-hand side (LHS) of the equation and use the double-angle formula for tangent.
LHS: cot(x) + tan(2x)
Using the double-angle formula for tangent: tan(2x) = 2tan(x) / (1 - tan^2(x))
Now substitute this in the equation:
LHS: cot(x) + (2tan(x) / (1 - tan^2(x)))
Next, we'll manipulate the right-hand side (RHS) of the equation using the reciprocal identities.
RHS: cot(x)sec^2(x)
Using the reciprocal identity: sec(x) = 1/cos(x) and sec^2(x) = 1/cos^2(x)
RHS: cot(x) * (1/cos^2(x))
Now, we'll cancel out common factors and simplify both sides of the equation.
LHS: cot(x) + (2tan(x) / (1 - tan^2(x)))
= cot(x) + (2tan(x) / (sec^2(x) - tan^2(x)))
Since sec^2(x) - tan^2(x) is identical to 1, we can further simplify:
LHS: cot(x) + (2tan(x) / 1)
= cot(x) + 2tan(x)
Comparing this to the RHS: cot(x) * (1/cos^2(x))
If you'd like, we can simplify the RHS as well:
RHS: cot(x) * (1/cos^2(x))
= cot(x)/cos^2(x)
Since cot(x) can be rewritten as cos(x)/sin(x):
RHS: (cos(x)/sin(x)) / cos^2(x)
= cos(x)/sin(x) * 1/cos^2(x)
= cos(x)/cos^2(x)sin(x)
= 1/(cos(x)sin(x))
= 1/(sin(x)/cos(x))
= cos(x)/sin(x)
Now, we can see that the LHS is equal to the RHS:
LHS: cot(x) + 2tan(x)
RHS: cos(x)/sin(x)
Therefore, we have proved that cot(x) + tan(2x) = cot(x)sec^2(x).