Question

prove the identity by use of the following fundamental identities.
csc^2ytan^2y=tan^2y+1

Answer 1

LHS
=1/ sin^y . sin^2 y/cos^2y
=1/cos^2y
=(sin^2y+cos^2y)/cos^y
=tan^2y+1
=RHS

Answer 2

csc^2 y tan^2 y - tan^2 y = 1
tan^2 y (csc^2 y - 1) = 1
csc^2 y - 1 = cot ^2 y
which is a pythagorean identity :)

Answer 3

so the answer here is correct http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet_Reduced.pdf

Answer 4

what is rhs? haha

Answer 5

Right hand side lol !!

Answer 6

can your solution will be the answer?

Answer 7

yes !

Answer 8

thx!

Answer 9

another 1 i post it hehe

Answer 10

(cotx+1)^2=csc^2x+2cotx

Answer 11

LHS
cot^2x+2cotx+1
since 1+cot^2x=csc^2x
csc^2x+2cotx
RHS

Answer 12

what is rhs?

Answer 13

LHS - Left Hand Side
RHS - Right Hand Side

Answer 14

since LHS=RHS,
instead of writing the term i just put LHS

Answer 15

ic ic thx

Answer 16

so what is the purpose of rhs in my question?

Answer 17

u have given both LHS term RHS term, u want to prove how LHS term equals RHS term.is n't it?

Answer 18

ahhh i get it now

Answer 19

but if i write it on my h.w.will i put it or it is ok that rhs/lhs is not there?

Answer 20

In some books, the proof of any such things is given in LHS/RHS form. But for ur case better ask ur teacher.

Answer 21

last 1 sir \[{1-\cos ^{2}\beta}\over {\cos \beta} \] =sin B tan B

Answer 22

LHS
=((sin^2B+cos^2B)-cos^2B)/cosB
=sin^2B/cosB
=sinB.sinB/cosB
=sinB tanB
RHS

Answer 23

thx sir your the best last pls (cscx-cotx) (cscx+cotx)=1

Answer 24

LHS
=csc^2x-cot^2x
=1
RHS

Answer 25

thx again gowthaman

Answer 26

(sin^2+cos^2)=1+2sin2cos2

Answer 27

something wrong

Answer 28

yes theres is w8

Answer 29

(sin^2+cos^2)^2=1+2sin2cos2

Answer 30

still something is wrong
it should be (sinx + cosx)^2 = 1 + 2sinx cosx or = 1+sin2x

Answer 31

i thought its 2 sory again >>> here (sinx + cosx)^2 = 1 + 2sinx cosx

Answer 32

expand it like (a+b)^2=a^2+b^2+2ab

Answer 33

in the last part is that 2sinxcosx?

Answer 34

2AB=2SINXCOSX?

Answer 35

S

Answer 36

s?

Answer 37

YES

Answer 38

sory lag thx gowthaman

Answer 39

my sister have this homework do you know some of this?

obtain the positive solution less than 360 degree
1. sec^2 x-4=0

2. cot^2x=3

3. 2cos^2x=cosx

4. cot^2x=cotx

5. 4sin^2x-3=0

Answer 40

its like some trigonometry angle or something quadrant

Answer 41

1. sec^2x-4=0
sec^2x=4
secx=+/- 2
or cosx = +/- 1/2
I think now u can solve for x