Expert answers to "Solve the equation 1+tan2A1+cot2A=(1−tan⁡A1−cot⁡A)2=tan2A"

High School
Geometry
Trigonometry
Trigonometric equations and identities

Bergen

Answered question

2020-10-31

Solve the equation $\frac{1 + \tan^{2} A}{1 + \cot^{2} A} = {(\frac{1 - \tan A}{1 - \cot A})}^{2} = \tan^{2} A$

Answer & Explanation

tafzijdeq

Skilled2020-11-01Added 92 answers

Work with the left side
$\frac{1 + \tan^{2} A}{1 + \cot^{2} A}$
$= \frac{1 + (\frac{\sin^{2} A}{\cos^{2} A})}{1 + (\frac{\cos^{2} A}{\sin^{2} A})} [∵ \tan A = \frac{\sin A}{\cos A}]$
$= \frac{\frac{\cos^{2} A + \sin^{2} A}{\cos^{2} A}}{\frac{\sin^{2} A + \cos^{2} A}{\sin^{2} A}}$
$= \frac{\frac{1}{\cos^{2} A}}{\frac{1}{\sin^{2} A}} [∵ \sin^{2} A + \cos^{2} A = 1]$
$= \frac{\sin^{2} A}{\cos^{2} A} = \tan^{2} A$
Now note that
$((1 - \tan A) / (1 - \cot A))^{2}$
$= {(\frac{1 - \frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}})}^{2} [∵ \tan A = \frac{\sin A}{\cos A}]$
$= {(\frac{\frac{\cos A - \sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}})}^{2}$
$= ((- (s i n A - c o s A)) / c o s A * s i n A / (s i n A - c o s A))^{2}$
$= {(- \frac{\sin A}{\cos A})}^{2} = \frac{\sin^{2} A}{\cos^{2} A} = \tan^{2} A$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Trigonometry

Didn't find what you were looking for?