How do you prove #(1+tany)/(1+coty)=secy/cscy#?

Question

Caroline Adams · Accepted Answer

See below.

Apply the following identities: •#tantheta = sintheta/costheta# •#cottheta = 1/tantheta = 1/(sintheta/costheta) = costheta/sintheta# •#sectheta = 1/costheta# •#csctheta= 1/sintheta# Start the simplification process on both sides. #(1 + siny/cosy)/(1 + cosy/siny) = (1/cosy)/(1/siny)# Put on a common denominator: #((cosy + siny)/cosy)/((siny + cosy)/siny) = 1/cosy xx siny/1# #(cosy + siny)/cosy xx siny/(siny + cosy) = siny/cosy# #(cancel(cosy + siny))/cosy xx siny/(cancel(siny + cosy)) = siny/cosy# #siny/cosy = siny/cosy# Identity proved!! Hopefully this helps!

Jordan Armstrong · Answer

undefined To prove $ \frac{1 + \tan y}{1 + \cot y} = \frac{\sec y}{\csc y} $, we can start with the left-hand side (LHS) and manipulate it step by step:

$$ \frac{1 + \tan y}{1 + \cot y} $$

Using the identities $ \tan y = \frac{\sin y}{\cos y} $ and $ \cot y = \frac{\cos y}{\sin y} $, we substitute these into the expression:

$$ \frac{1 + \frac{\sin y}{\cos y}}{1 + \frac{\cos y}{\sin y}} $$

Next, we simplify the fractions by multiplying each term by the respective denominator:

$$ \frac{\cos y + \sin y}{\cos y + \sin y} $$

Since the numerator and denominator are the same, they cancel out, leaving:

$$ 1 $$

Now, let's simplify the right-hand side (RHS) of the equation:

$$ \frac{\sec y}{\csc y} $$

Using the reciprocal identities $ \sec y = \frac{1}{\cos y} $ and $ \csc y = \frac{1}{\sin y} $, we substitute these into the expression:

$$ \frac{\frac{1}{\cos y}}{\frac{1}{\sin y}} $$

Simplifying by multiplying by the reciprocal of the denominator:

$$ \frac{\sin y}{\cos y} $$

Finally, since $ \frac{\sin y}{\cos y} = \tan y $, we have:

$$ \frac{\sec y}{\csc y} = \tan y $$

Therefore, we have shown that the LHS equals the RHS, proving the given identity.