Question

any one wanna !! learn trigonometry or wanna revise it!! I've some ques

Answer 1

reply u r levels also

Answer 2

i am just here to revise

Answer 3

oh!
@UnkleRhaukus

here's a ques... for u!
GIVEN:
cotA+tanA = x
& secA - cosA = y....
then which of the following is true ?
(a) (xy^2)^2/3 - (x^2y)^2/3 = 1
(b) (x^2y)^2/3 - (xy^2)^2/3 = 1
(c) x^2/3+y^2/3 = 1
(d) none of these
EXPLAIN AND WRITE EACH STEP AND CONCEPT.....

Answer 4

can you give me a clue

Answer 5

clue!- difficult trig. ques.

Answer 6

do u want the solution

Answer 7

i want you to teach me how to find the solution myself

Answer 8

hi, lets test this by putting a real number. A=30 degrees. (because for 30 degree we've all values defined in the table)
then cotA+tanA = x =4/root 3
and secA - cosA = y=1/(2*root 3)

now you put these values of x and y in LHS of first three equations, I'm not getting RHS=1.
that's why right answer is (d) none of this.

Answer 9

@UnkleRhaukus ^^

Answer 10

that is not a mathematical solution

Answer 11

ya its ! logically its!

Answer 12

@rs32623 u r still writing!

Answer 13

it will go like this
cotA+tanA = x
& secA - cosA = y....

squaring both & subtracting 1st from 2nd we have
-3+(cos^2a-cot^2a)=y^2-x^2

Now we have from 1st equation cota-tana=sqrt(x^2-4)
& from 2nd we have seca+cosa=sqrt(y^2+4)

From these two new equation we can find cos^2a & cot^2a in terms of x & y.
Now put these values in equation & find the expression.
I am calculating right now & report it in few minutes

Answer 14

oh

Answer 15

smells like none of these. may be i've mistaken at calculation but steps are perfectly correct you can give it a try.

I used the concept that (a-b)=\[\sqrt{(a+b)^{2}-4ab}\]
& (a+b)=\[\sqrt{(a-b)^{2}+4ab}\]
From these we can find the cosa & hence the cos^2a by subtracting 2 from 5 similarly cota & hence cot^2a by adding 4 & 1

Also Sec^2a-tan^2=1 & seca*cosa=1 & cot*tana=1