Question

Tg^2alpha-sin^2alpha=tg^2alpha sin^2alpha

Answer 1

Tg²(α) - sin²(α) = tg²(α)sin²(α)
sin²(α)/cos²(α) - sin²(α) = tg²(α)sin²(α)
делим обе части уравнения на sin²(α)
1/cos²(α) - 1 = tg²(α)
(1 - cos²(α))/cos²(α) = sin²(α)/cos²(α)
умножаем обе части уравнения на cos²(α)
1 - cos²(α) = sin²(α)
1 = sin²(α) + cos²(α)
1 = 1 
Верное тождество.

Answer 2

To prove the trigonometric identity $\tan^2\alpha - \sin^2\alpha = \tan^2\alpha \sin^2\alpha$, we'll manipulate the left-hand side (LHS) and right-hand side (RHS) separately and show that they are equal.

Starting with the LHS: $\tan^2\alpha - \sin^2\alpha$

We know the Pythagorean Identity for tangent: $\tan^2\alpha = \sec^2\alpha - 1$

So we can rewrite the LHS as: $\sec^2\alpha - 1 - \sin^2\alpha$

Next, we'll use the Pythagorean Identity for sine: $\sin^2\alpha = 1 - \cos^2\alpha$

Substitute this into the LHS: $\sec^2\alpha - 1 - (1 - \cos^2\alpha)$

Simplify: $\sec^2\alpha - 1 - 1 + \cos^2\alpha$

Combine like terms: $\sec^2\alpha - 2 + \cos^2\alpha$

Now, we'll use the Pythagorean Identity for cosine: $\cos^2\alpha = 1 - \sin^2\alpha$

Substitute this into the LHS: $\sec^2\alpha - 2 + (1 - \sin^2\alpha)$

Simplify: $\sec^2\alpha - 1 - \sin^2\alpha$

Now, this matches the original LHS expression: $\tan^2\alpha - \sin^2\alpha$

Moving on to the RHS: $\tan^2\alpha \sin^2\alpha$

Now, we know the tangent-sine identity: $\tan\alpha \sin\alpha = \sin^2\alpha$

So, we can rewrite the RHS as: $\tan^2\alpha \sin^2\alpha = \tan^2\alpha \cdot (\tan\alpha \sin\alpha)$

Using the tangent-sine identity, we can replace $\tan\alpha \sin\alpha$ with $\sin^2\alpha$: $\tan^2\alpha \cdot \sin^2\alpha = \tan^2\alpha \cdot \sin^2\alpha$

We can see that the RHS matches the original expression: $\tan^2\alpha \sin^2\alpha$

Since the LHS and RHS are both equal to $\tan^2\alpha \sin^2\alpha$, we have proven the identity: $\tan^2\alpha - \sin^2\alpha = \tan^2\alpha \sin^2\alpha$