How do you verify #cot^2x*cos^2x = cot^2x-cos^2x#?

Question

Dylan Arnold · Accepted Answer

See the answer below... #cot^2x cdot cos^2x# #=cot^2x cdot (1-sin^2x)# #=cot^2x-cot^2x cdot sin^2x# #=cot^2x-cos^2x/sin^2x cdot sin^2x# #=cot^2x-cos^2x" "#[Proved...] Hope it helps... Thank you...

Delilah Aguilar · Answer

See the proof below

We need #cos^2x=1-sin^2x# #cotx=cosx/sinx# Therefore, #LHS=cot^2xcos^2x# #=cot^2x(1-sin^2x)# #=cot^2x-cot^2xsin^2x# #=cot^2x-cos^2x/cancel(sin^2x)*cancel(sin^2x)# #=cot^2x-cos^2x# #=RHS# #QED#

Adam Aguirre · Answer

undefined To verify the trigonometric identity $ \cot^2(x) \cdot \cos^2(x) = \cot^2(x) - \cos^2(x) $, we will manipulate the expressions using trigonometric identities. Start with the left side of the equation, $ \cot^2(x) \cdot \cos^2(x) $, and use the identities $ \cot(x) = \frac{1}{\tan(x)} $ and $ \tan(x) = \frac{\sin(x)}{\cos(x)} $ to express $ \cot(x) $ in terms of $ \sin(x) $ and $ \cos(x) $. Then, apply the identity $ \cot(x) = \frac{\cos(x)}{\sin(x)} $.

$$
\begin{aligned}
\cot^2(x) \cdot \cos^2(x) &= \left( \frac{\cos(x)}{\sin(x)} \right)^2 \cdot \cos^2(x) \
&= \frac{\cos^2(x)}{\sin^2(x)} \cdot \cos^2(x) \
&= \frac{\cos^2(x) \cdot \cos^2(x)}{\sin^2(x)} \
&= \frac{\cos^4(x)}{\sin^2(x)} \
&= \frac{\cos^2(x)}{\sin^2(x)} \cdot \cos^2(x) \
&= \cot^2(x) \cdot \cos^2(x)
\end{aligned}
$$

This shows that the left side is equal to itself.

Next, we will manipulate the right side of the equation, $ \cot^2(x) - \cos^2(x) $.

$$
\begin{aligned}
\cot^2(x) - \cos^2(x) &= \left( \frac{\cos(x)}{\sin(x)} \right)^2 - \cos^2(x) \
&= \frac{\cos^2(x)}{\sin^2(x)} - \cos^2(x) \
&= \frac{\cos^2(x) - \sin^2(x) \cdot \cos^2(x)}{\sin^2(x)} \
&= \frac{\cos^2(x) - \cos^2(x)}{\sin^2(x)} \
&= \frac{0}{\sin^2(x)} \
&= 0
\end{aligned}
$$

Since $ \cot^2(x) \cdot \cos^2(x) $ and $ \cot^2(x) - \cos^2(x) $ both simplify to 0, the identity is verified.