Question

Prove that the equation is an identity. \frac{\tan A - 1}{\tan A + 1} = \frac{1 - \cot A}{1 + \cot A} Working from the left-hand side, multiplying the numerator and denominator by \cot A. LHS = \frac{\tan A - 1}{\tan A + 1} = \frac{(\tan A - 1)}{(\tan A + 1)} = \frac{\tan A - }{\tan A + } Use the relationship between \tan x and \cot x to simplify. \frac{\tan A - 1}{\tan A + 1} =

          Prove that the equation is an identity.
\frac{\tan A - 1}{\tan A + 1} = \frac{1 - \cot A}{1 + \cot A}
Working from the left-hand side, multiplying the numerator and denominator by \cot A.
LHS = \frac{\tan A - 1}{\tan A + 1}
= \frac{(\tan A - 1)}{(\tan A + 1)}
= \frac{\tan A - }{\tan A + }
Use the relationship between \tan x and \cot x to simplify.
\frac{\tan A - 1}{\tan A + 1} =

Prove that the equation is an identity.
(tanA - 1)/(tanA + 1) = (1 - cotA)/(1 + cotA)
Working from the left-hand side, multiplying the numerator and denominator by cotA.
LHS = (tanA - 1)/(tanA + 1)
= ((tanA - 1))/((tanA + 1))
= (tanA - )/(tanA + )
Use the relationship between tanx and cotx to simplify.
(tanA - 1)/(tanA + 1) =

Added by Nicole M.

Introductory and Intermediate Algebra for College Students 4th

Robert Blitzer 4th Edition

Instant Answer

Solved by Expert Steven Clarke

06/19/2022

Step 1

Step 1: Start with the given equation: (tanA-1)/(tanA+1) = (1-cotA)/(1+cotA) Show more…

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Prove that the equation is an identity: (tanA-1)/(tanA+1) = (1-cotA)/(1+cotA) Working from the left-hand side, multiplying the numerator and denominator by cotA: LHS = (tanA-1)/(tanA+1) Use the relationship between tanx and cotx to simplify: (tanA-1)/(tanA+1) = (tanA-1)/(tanA+1) Need Help? Prove that the equation is an identity: tanA-1 = 1-cotA tanA+1 = 1+cotA Working from the left-hand side, multiplying the numerator and denominator by cotA: tanA+1 = cotA(tanA+1) tanA-1 = cotA(tanA-1) Use the relationship between tanx and cotx to simplify: tanA-1 = cotA(tanA-1) tanA+1 = cotA(tanA+1) Need Help? ReadIt Submit Answer