Question

(cos^2t - ctg^2t)/(sin^2t - tg^2t)

Answer 1

$\frac{Cos^{2}t-Ctg^{2}t }{Sin{2}t-tg^{2}t}=\frac{Cos^{2}t-\frac{Cos^{2}t }{Sin^{2}t}}{Sin^{2}t-\frac{Sin^{2}t }{Cos^{2}t }} =\frac{\frac{Sin^{2}tCos^{2}t-Cos^{2}t}{Sin^{2}t}}{\frac{Sin^{2}tCos^{2}t-Sin^{2}t}{Cos^{2}t }}=\frac{Cos^{2}t(Sin^{2}t-1)*Cos^{2}t }{Sin^{2}t(Cos^{2}t-1)*Sin^{2}t}=\frac{-Cos^{2}t*Cos^{4}t}{-Sin^{2}t*Sin^{4}t}=\frac{Cos^{6}t }{Sin^{6}t }=Ctg^{6}t$

Answer 2

To simplify the given expression, let's use trigonometric identities to manipulate it. First, we'll write each trigonometric function in terms of sine and cosine:

1. cotangent (ctg t) is equal to 1/tan(t) = cos(t)/sin(t).
2. tangent (tg t) is equal to sin(t)/cos(t).

Now, let's rewrite the expression using these identities:

(cos^2(t) - (cos(t)/sin(t))^2) / (sin^2(t) - (sin(t)/cos(t))^2)

Next, simplify further:

(cos^2(t) - cos^2(t)/sin^2(t)) / (sin^2(t) - sin^2(t)/cos^2(t))

To get a common denominator, multiply the numerator and denominator of the first fraction by sin^2(t), and the second fraction by cos^2(t):

(sin^2(t)*cos^2(t) - cos^2(t)) / (sin^2(t)*cos^2(t) - sin^2(t))

Now, factor out the common term from both the numerator and the denominator:

cos^2(t) * (sin^2(t) - 1) / (sin^2(t) * (cos^2(t) - 1))

Next, use the trigonometric identity sin^2(t) + cos^2(t) = 1:

cos^2(t) * (-cos^2(t)) / (sin^2(t) * (-sin^2(t)))

Now, cancel out the common negative term:

cos^2(t) / sin^2(t)

Finally, use the identity tan^2(t) = sin^2(t)/cos^2(t):

cos^2(t) / (tan^2(t) * cos^2(t))

Now, we can cancel out the common factor cos^2(t):

1 / tan^2(t)

So, the simplified form of the expression is 1 / tan^2(t).