+9481 \(\dfrac{1+\cos2\theta}{\sin2\theta}=\cot\theta\)
Let's turn the left side into the right side.
\(\phantom{=\,}\ \dfrac{1+\cos2\theta}{\sin2\theta}\)
First let's use the double-angle identity for sin: \(\sin2θ=2 \sinθ \cosθ\)
\(=\,\dfrac{1+\cos2\theta}{2\sin\theta\cos\theta}\)
Now let's use this double-angle identity for cos: \(\cos2\theta=2\cos^2\theta-1\)
\(=\,\dfrac{1+2\cos^2\theta-1}{2\sin\theta\cos\theta}\)
Add together the 1 and the -1 in the numerator to get 0 .
\(=\,\dfrac{2\cos^2\theta}{2\sin\theta\cos\theta}\)
Divide the numerator and denominator by 2 .
\(=\,\dfrac{\cos^2\theta}{\sin\theta\cos\theta}\)
Divide the numerator and denominator by \(\cos\theta\) .
\(=\,\dfrac{\cos\theta}{\sin\theta}\)
And by the quotient identity for cotangent, \(\cot\theta=\frac{\cos\theta}{\sin\theta}\)
\(=\,\cot\theta\)
