Question

Ctg a-cos2a•ctg a=sin2 a

Answer 1

$Ctg\alpha-Cos2\alpha*Ctg\alpha=Ctg\alpha (1-Cos2\alpha)=\\\\=Ctg\alpha[Sin^{2}\alpha+Cos^{2}\alpha -(Cos^{2}\alpha-Sin^{2}\alpha)]=\\\\=Ctg\alpha(Sin^{2}\alpha +Cos^{2}\alpha-Cos^{2} \alpha+Sin^{2}\alpha)=\frac{Cos\alpha}{Sin\alpha}*2Sin^{2}\alpha=\\\\=2Sin\alpha Cos\alpha=Sin2\alpha\\\\Sin2\alpha =Sin2\alpha$

Answer 2

To prove the trigonometric identity $\cot(a) - \cos^2(a) \cdot \cot(a) = \sin^2(a)$, we'll start with the left-hand side (LHS) and manipulate it step by step to hopefully arrive at the right-hand side (RHS):

Starting with LHS: $\cot(a) - \cos^2(a) \cdot \cot(a)$

We'll factor out $\cot(a)$ from both terms: $\cot(a) \cdot (1 - \cos^2(a))$

We know that $\sin^2(a) = 1 - \cos^2(a)$, which is a Pythagorean identity for sine and cosine. Substituting this in: $\cot(a) \cdot \sin^2(a)$

Now we can see that the expression on the left-hand side matches the right-hand side $\sin^2(a)$. Thus, the identity is proven:

LHS: $\cot(a) - \cos^2(a) \cdot \cot(a) = \cot(a) \cdot (1 - \cos^2(a)) = \cot(a) \cdot \sin^2(a) = \sin^2(a)$

RHS: $\sin^2(a)$

Therefore, the given trigonometric identity $\cot(a) - \cos^2(a) \cdot \cot(a) = \sin^2(a)$ is indeed true.