Question

Sin^2a/tg^2a + cos^2a/ctg^2a- 1/sin^2a=-ctg^2a​

Answer 1

$\frac{Sin^{2}\alpha}{tg^{2}\alpha}+\frac{Cos^{2}\alpha}{Ctg^{2}\alpha}-\frac{1}{Sin^{2}\alpha}=\frac{Sin^{2}\alpha Cos^{2}\alpha}{Sin^{2}\alpha}+\frac{Cos^{2}\alpha Sin^{2}\alpha}{Cos^{2}\alpha}-\frac{1}{Sin^{2}\alpha} =Cos^{2}\alpha+Sin^{2}\alpha-\frac{1}{Sin^{2}\alpha}=1-\frac{1}{Sin^{2}\alpha} =\frac{Sin^{2}\alpha-1}{Sin^{2}\alpha}=\frac{-Cos^{2}\alpha}{Sin^{2}\alpha}=-Ctg^{2} \alpha\\\\-Ctg^{2}\alpha=-Ctg^{2}\alpha$

Тождество доказано

Answer 2

To prove the given trigonometric identity, we will manipulate the left-hand side (LHS) of the equation step by step:

LHS = sin^2(a)/tan^2(a) + cos^2(a)/cot^2(a) - 1/sin^2(a)

To simplify the expression, we'll use the trigonometric identity: cot^2(a) = 1/tan^2(a)

Substitute cot^2(a) in the second term:

LHS = sin^2(a)/tan^2(a) + cos^2(a)/(1/tan^2(a)) - 1/sin^2(a)

Now, to add the fractions in the first two terms, we need a common denominator. The common denominator for tan^2(a) and 1/tan^2(a) is tan^2(a):

LHS = (sin^2(a) + cos^2(a)*tan^2(a) - 1)/tan^2(a)

Now, remember the Pythagorean identity: sin^2(a) + cos^2(a) = 1

Substitute this into the expression:

LHS = (1*tan^2(a) - 1)/tan^2(a)

Now, simplify further:

LHS = (tan^2(a) - 1)/tan^2(a)

Next, we can use the identity: tan^2(a) = 1 + cot^2(a)

Substitute this into the expression:

LHS = [(1 + cot^2(a)) - 1]/(1 + cot^2(a))

Simplify:

LHS = cot^2(a)/ (1 + cot^2(a))

Now, recall the identity: cot^2(a) = 1/tan^2(a)

Substitute this back in:

LHS = (1/tan^2(a))/(1 + 1/tan^2(a))

LHS = 1/(tan^2(a) + 1/tan^2(a))

Now, recall the identity: tan^2(a) + 1/tan^2(a) = sec^2(a)

Substitute this into the expression:

LHS = 1/sec^2(a)

Finally, remember the definition of secant: sec(a) = 1/cos(a)

Substitute this in:

LHS = 1/(1/cos^2(a))

LHS = cos^2(a)

Thus, we have proved that:

LHS = cos^2(a) = -cot^2(a)

So, the given trigonometric identity is true.