Question

Simplify the trigonometric expression.

Answer 1

@kola908

Answer 2

do you have your list of trig identities?

Answer 3

No idea how to do this! sorry

Answer 4

let me see you try to solve without thinking that sine is sine. think of it as regular x and see where that will take you

Answer 5

\[\frac{ 1 }{ 1+x }+\frac{ 1 }{ 1-x }\]
can you solve this?

Answer 6

I told you I have no idea how to do this.. sorry..

Answer 7

hmm can you solve this ?\[(1-x) (1+x) \]

Answer 8

1x^2

Answer 9

?

Answer 10

"FOIL" method

Answer 11

\[(A+B) (A-B) = A*A -A*B +A*B+B*B \]
simplifies to \[A^2+B^2\]

Answer 12

ok

Answer 13

\[1+\sin^2\theta \]
right?

Answer 14

are you making any connection?

Answer 15

means 1

Answer 16

I think you need help way beyond trigonometry :(

Answer 17

going back to introduction to algebra

Answer 18

@nincompoop

I'm sure you meant \(\large (A + B)(A - B) = A^2 \color{red}- B^2\)

and

\( \large (1 + \sin \theta)(1 - \sin \theta) = 1 \color{red}- \sin^2 \theta\)

Answer 19

uyeah
thanks for correcting

Answer 20

yw

Answer 21

can someone just please show me step by step please

Answer 22

You are adding two fractions. That means you need a common denominator.

Answer 23

The LCD is \( (1 + \sin \theta)(1 - \sin \theta) \)

Answer 24

\[1-\sin^2 \theta = \cos^2 \theta\]

Answer 25

We need to multiply the left fraction by \( \dfrac{1 - \sin \theta}{1 - \sin \theta}\)

and the right fraction by \( \dfrac{1 + \sin \theta}{1 + \sin \theta}\)

Answer 26

see
\[\frac{ 1 }{ \cos \theta } = \sec \theta\]

you don't have to know a lot
you just need to make sure you can start deducing the possible answers

Answer 27

so this means that our answer contains
\[\sec^2 \theta\]

Answer 28

A... 2 Sec^3 O

Answer 29

it can't have an order magnitude of 3 because our denominator has only an order of 2

Answer 30

well its the only answer with sec

Answer 31

Please this is my last question...

Answer 32

i meant 2 sec 2... look at A

Answer 33

\(\dfrac{1}{1 + \sin \theta} + \dfrac{1}{1 - \sin \theta} \)

\(= \dfrac{1}{1 + \sin \theta} \times \dfrac{1 - \sin \theta}{1 - \sin \theta} + \dfrac{1}{1 - \sin \theta} \times \dfrac{1 + \sin \theta}{1 + \sin \theta}\)

\(= \dfrac {1 - \sin \theta} {(1 + \sin \theta)(1 - \sin \theta) } + \dfrac{1 + \sin \theta}{(1 + \sin \theta)(1 - \sin \theta)}\)

Answer 34

Im choosing A

Answer 35

you are correct
but you need to know the techniques

Answer 36

\(= \dfrac {1 - \sin \theta} {1 - \sin^2 \theta } + \dfrac{1 1 + + \sin \theta}{1 - \sin^2 \theta}\)

\(= \dfrac {1 - \sin \theta +1 - \sin^2 \theta }{1 - \sin^2 \theta}\)

\(= \dfrac {2}{\cos^2 \theta}\)

\(= 2 \sec^2 \theta\)