Question

tanA=x , sec2A = ?
a. (1 + x^2)/(1 - x^2)
b. (1-x^2)/2x

Answer 1

a)

Answer 2

how did u get that?? pls tell

Answer 3

Sec2A=(1+tan^2A)/(1-tan^2A) so Sec2A=1 + x^2)/(1 - x^2)

Answer 4

okie...thank u.. :)

Answer 5

My pleasure........

Answer 6

can u pls help me once more??

Answer 7

Allrite just remember the formula for cos 2A=(1-tan^2A)/(1+tan^2A)
Sec 2A=1/cos 2A So, Sec 2A=(1+tan^2A)/(1-tan^2A)
Now Substitute for tan A

Answer 8

k....another qstn ?

Answer 9

if tan(πcos\[\theta\] = cot(πsin\[\theta\] , then cos(\[\theta\]- π/4) =

Answer 10

Couldnot understand the question, pls type it properly so that I can help you

Answer 11

problem with equation editor.. :(
tan(πcos theta)= cot (π sin theta) , then cos(theta - π/4) =

Answer 12

a. 2
b. 1/2√2

Answer 13

Are you sure that the question is correct!

Answer 14

yes...it θ

Answer 15

i mean πcosθ

Answer 16

It is b) but I am still working on it

Answer 17

thanks 4 helping me...how did u get b?

Answer 18

tan(a)=x/1=opp/adj
sec(2a)=1/[cos(2a)]=1/[cos^2a-sin^2a]=1/[(cosa)^2-(sina)^2]
hyp=sqrt{opp^2+adj^2}=sqrt{x^2+1}

cosa=adj/hyp=1/[sqrt{x^2+1}]

(cosa)^2=1/(x^2+1)

sina=opp/hyp=x/[sqrt{x^2+1}]

(sina)^2=x^2/[x^2+1]

(cosa)^2-(sina)^2=(1-x^2)/(x^2+1)

sec(2a)=(x^2+1)/(1-x^2)

Answer 19

thank u .. :) can u pls help me with the other one too?

Answer 20

ok im trying to get ready for something
if i have time i will answer

Answer 21

okie..=D