Question

Verify this:

Answer 1

\[(\tan \theta + \sec \theta-1) /(\tan \theta - \sec \theta + 1) = 1 + \sin \theta / \cos \theta\]

Answer 2

@gorv

Answer 3

rewrite the LHS using only sin and cos

Answer 4

thats what I'm trying to do and I can't figure it out

Answer 5

like sin/cos? and 1/cos?

Answer 6

right

Answer 7

ok then what?

Answer 8

\[LHS\\={\tan \theta +\sec \theta -1 \over \tan \theta -\sec \theta+1}\\ \large=\color{blue}{{\sin \theta \over \cos \theta} +{1 \over \cos \theta}-1 \over {\sin \theta \over \cos \theta} -{1 \over \cos \theta}+1 }\]

Answer 9

yes that is what i got

Answer 10

then do yo do common denominators for both sides

Answer 11

you*

Answer 12

how about we next multiply the numerator and denom by cos(theta)

Answer 13

i meant both top and bottom

Answer 14

ok so you get \[\sin \theta - \cos \theta +1 / \sin \theta + \cos \theta - 1\]

Answer 15

you don't have to spend so much time writing the equation you can just explain it to me

Answer 16

\[={\sin \theta+1- \cos \theta \over \sin \theta-1+ \cos \theta}\\={\sin \theta- (\cos \theta-1) \over \sin \theta+ (\cos \theta-1)}\]

maybe multiplying the numerator to both the numerator and denom would help next

Answer 17

what would the numerator be to multiply, that whole expression?\[\sin \theta - (\cos \theta - 1)\]

Answer 18

yes ...i am thinking that would help

Answer 19

i don't know what the result of multiplying the numerator would be

Answer 20

ok so i got
\[\sin ^2 \theta - 2\sin \theta \cos \theta + 2 \sin \theta + ( \cos \theta - 1) ^2 / \sin ^2 \theta - ( \cos \theta - 1) ^2\]

Answer 21

\[\color{blue}{\sin^2\theta} -2\sin \theta(\cos \theta -1)\color{blue}{+\cos^2\theta}-2\cos \theta+1 \over \color{blue}{\sin^2\theta} - \cos^2 \theta+2\cos \theta \color{blue}{-1}\]

Answer 22

ok thats what i got I guess and then what

Answer 23

replace what i marked in blue

sin^2+cos^2=1

sin^2-1= -cos^2

Answer 24

ok and then what

Answer 25

\[=-{2\sin \theta(\cos \theta-1)-2\cos+2 \over -2\cos^2\theta+2\cos \theta}\]

looks like numerator and denom have something in common ... reduce

Answer 26

**\[={\color{red}-2\sin \theta(\cos \theta-1)-2\cos+2 \over -2\cos^2\theta+2\cos \theta}\]

Answer 27

ok what is the reduced form?

Answer 28

\[={-2\left( \cos \theta-1 \right)\left[ \sin \theta+1 \right] \over -2\left( \cos \theta-1 \right)\cos \theta}\]

Answer 29

thanks