How do you prove #(cscx + cotx)^2 = (1 + cosx) / (1 - cosx)#?

Question

Oliver Atkinson · Accepted Answer

in this manner. The initial participant is: #(1/sinx+cosx/sinx)^2=(1+cosx)^2/sin^2x=(1+cosx)^2/(1-cos^2x)=# #(1+cosx)^2/((1+cosx)(1-cosx))=(1+cosx)/(1-cosx)#, That makes up the second participant.

Daniel Acosta · Answer

#(csc x + cotx)^2# = #(1/sin x + cos x/ sin x)^2# = # (1+cosx)^2 / sin^2 x# = #(1+cos x)^2 /(1-cos^2x)# =#(1+cosx)^2 /((1+cos x)(1-cos x))# = # (1+cos x)/(1-cosx)#  = RHS

Delilah Aguilar · Answer

undefined To prove the trigonometric identity $ (\csc x + \cot x)^2 = \frac{1 + \cos x}{1 - \cos x} $, follow these steps:

Start with the left-hand side (LHS) of the equation:
$$ (\csc x + \cot x)^2 $$

Expand and simplify the expression:
$$ (\csc x + \cot x)^2 = \csc^2 x + 2\csc x \cot x + \cot^2 x $$

Use the trigonometric identities:
$$ \csc^2 x = \frac{1}{\sin^2 x} $$
$$ \cot x = \frac{\cos x}{\sin x} $$
$$ \cot^2 x = \left( \frac{\cos x}{\sin x} \right)^2 = \frac{\cos^2 x}{\sin^2 x} $$

Substitute these identities into the expression:
$$ \frac{1}{\sin^2 x} + 2\frac{\cos x}{\sin x} \cdot \frac{\cos x}{\sin x} + \frac{\cos^2 x}{\sin^2 x} $$

Combine the fractions:
$$ \frac{1}{\sin^2 x} + \frac{2\cos^2 x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} $$

Combine the terms:
$$ \frac{1 + 3\cos^2 x}{\sin^2 x} $$

Now, use the Pythagorean identity $ 1 = \sin^2 x + \cos^2 x $ to simplify further:
$$ 1 + 3\cos^2 x = \sin^2 x + 3\cos^2 x = \sin^2 x + 3(1 - \sin^2 x) $$
$$ = \sin^2 x + 3 - 3\sin^2 x = 3 - 2\sin^2 x $$

Substitute this back into the expression:
$$ \frac{3 - 2\sin^2 x}{\sin^2 x} = \frac{3 - 2(1 - \cos^2 x)}{1 - \cos^2 x} $$
$$ = \frac{3 - 2 + 2\cos^2 x}{1 - \cos^2 x} = \frac{1 + 2\cos^2 x}{1 - \cos^2 x} $$

Since $ 1 - \cos^2 x = \sin^2 x $, the expression simplifies to:
$$ \frac{1 + \cos x}{1 - \cos x} $$

This is the same as the right-hand side (RHS) of the equation, so the identity is proven.