Question

Gud mornin evrybody.umm i need help in trignometry

Answer 1

oh another guy asked this too :D
http://openstudy.com/study#/updates/53c763b0e4b05c273e9e9a65


well do u know what is sec and cosec ?

Answer 2

yup

Answer 3

plsa write it step by step

Answer 4

ok,we'll solve it step by step.

Answer 5

kk

Answer 6

let's solve the first one...we can factor and simply write : 9(sec^2 A - tan^2 A)

Answer 7

and you know sec^2 A = 1/cos^2 A ... ok?

Answer 8

ya but y is 9 out of th ebracket

Answer 9

i factored...for example we have 9a + 9b ... we can factor it and write : 9(a+b)

Answer 10

ya since 9 is common

Answer 11

exactly :)
and sec^2 A = 1/cos^2A and as we have this:

\(\Large \color{Lime}{1 + \tan ^2 A = \frac{ 1 }{ \cos ^2 A }}\)

Answer 12

we can substite 1/cos^2A or sec^A

got it?

Answer 13

@plshelp9oup , are u confused?

Answer 14

yupppp

Answer 15

@PFEH.1999

Answer 16

you're confused?

ok ... why are u confused? what made u confused?

Answer 17

@plshelp9oup , where are u?

Answer 18

sry i have to go plz hurry up.

Answer 19

i cant understand wat u told @PFEH.1999

Answer 20

ok...did you get this that we can factor and make 9(sec^2 A - tan^2 A ) ?

Answer 21

hmm..

Answer 22

and have u seen this formula?

\( \Large \color{Lime}{1 + \tan ^2 A = \frac{ 1 }{ \cos ^2 A }} \)

Answer 23

ya

Answer 24

so as 1/cos^2 A = you can write :

\( \Large \color{Green}{1 + \tan ^2 A = sec^2 A } \)

Answer 25

1/cos^2A = sec^2 A

Answer 26

wht??

Answer 27

i said you that 1/cos^2 A = sec^2 A so we substite it and we'll be able to find that.

Answer 28

ya k

Answer 29

\( \color{Orange}{9 \sec^2 A - 9\tan^2 A = 9(\sec^2 A - \tan^2 A) = 9((1 + \tan^2A) - \tan^2 A) = 9(1) = 9} \)

Answer 30

umm whhic question ru doin

Answer 31

i am solving (i) (number one)

Answer 32

okay

Answer 33

got it?

Answer 34

leave dis quesiton bro lets do (iii)

Answer 35

ya nd i got it ( i mean the (i) i understood

Answer 36

ok,u can use the rule (a+b)(c-d) = ac + bc - ad - bd
:)

Answer 37

wht is tht rule is it same asa (a+b)(a-b) ??

Answer 38

u dere

Answer 39

no i meant that you can use this to solve number three.

Answer 40

you can write number three in this way:

\[\sec A + \tan A - (secA)(sinA) - (tanA)(sinA)\]

Answer 41

how did u do tht

Answer 42

i used this:

(a+b)(c-d) = ac + bc - ad - bd

Answer 43

this is an example which helps you

Answer 44

wait hold a sec dis is wht i did

Answer 45

ok,what was your final answer?

Answer 46

hold a sec plss

Answer 47

i don't see anything..please attach it

Answer 48

click the link

Answer 49

plz just tell me ur final answer.

Answer 50

the final answer is cos A

Answer 51

@plshelp9oup , do u want to know how?

Answer 52

i couldnt get it

Answer 53

ya it is cos a but how did u get it

Answer 54

try continuing frn where i stopped

Answer 55

well , i found

\( \sec A + \tan A - (secA)(sinA) - (tanA)(sinA) \)

but as (secA)(sinA) = sinA\sosA = tan A
so the expression would be:

\(\color{lime}{\sec A - (tanA)(sinA) }\)

Answer 56

sos A???

Answer 57

sos???

Answer 58

you mean cos A?

Answer 59

no u wrote sos a

Answer 60

oh sry ! i meant cos A

Answer 61

(secA)(sinA) = sinA\cosA = tan A

Answer 62

so we found \( \color{lime}{\sec A - (tanA)(sinA) } \)

but

\(\large (tanA)(sinA) = (\frac{ sinA }{ cosA })(sinA) = \frac{ \sin^2 A }{ cosA } \)

Answer 63

and do u know sin^2A + cos^2A = 1 ?

Answer 64

@plshelp9oup

Answer 65

you wait too much! what are u doing?

Answer 66

umm sorry i was eating

Answer 67

ok...i've to go..sry...i'll call one of my friends to continue helping u.

Answer 68

@ganeshie8

Answer 69

gee thnx

Answer 70

looks most of them are already solved,
which problem are you working on right now ? :)