Question

If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1

Answer 1

first of fall
sec theta=1/cos theta
and tan thata= sintheta/costheta
than you pu this value in rhs
from that you can proved that

Answer 2

u got it ?

Answer 3

No i don't understand what the last part means..

Answer 4

\[\large \sec(\theta)+ \tan(\theta) = P\] so \[\sin(\theta) = \frac{P^2-1}{P^2+1} \]

Answer 5

Thats the question?

Answer 6

lol., who's question is this?

Answer 7

ohk
1/cos + sin/cos=p
then you take the lcm
1+ sin/cos is the value of p then you put this value in p2-1/p2+1

Answer 8

Okay sonot trying to evaluate RHS but LHs....

Answer 9

you got it

Answer 10

what ?

Answer 11

but lhs..........

Answer 12

hey r u bsy

Answer 13

\[\sec(\theta) + \tan(\theta)\]\[\frac{1}{\cos(\theta)} +\frac{\sin(\theta)}{\cos(\theta)}\]\[\frac{1+\sin(\theta)}{\cos(\theta)}\] evaluating left hand side...but now what?

Answer 14

You can square the top and bottom, right?

Answer 15

u put this value in place of p
p2-1/p2+1

Answer 16

yup

Answer 17

ohk
bye bye
see you later

Answer 18

:|

Answer 19

hey friends bye bye

Answer 20

anyone have any doubt

Answer 21

i will be back soon

Answer 22

\[\frac{P^2-1}{P^2+1}\]\[=\frac{(sec\theta+tan\theta)^2-1}{(sec\theta+tan\theta)^2+1}\]\[=\frac{sec^2\theta+2sec\theta+tan^2\theta-1}{sec^2\theta+2sec\theta+tan^2\theta+1}\]\[=\frac{2sec\theta tan\theta+2tan^2\theta}{2sec^2\theta+2sec\theta tan\theta}\]\[=\frac{2tan\theta(sec\theta +tan\theta)}{2sec\theta(sec\theta+ tan\theta)}\]\[=sin\theta\]Can I use Q.E.D. here? :D

Answer 23

ohIsee now lol.

Answer 24

I got it xD

Answer 25

Sorry for the typos in the third line, it should be
\[=\frac{sec^2θ+2secθtanθ+tan^2θ−1}{sec^2θ+2secθtanθ+tan^2θ+1}\]
It can be re arranged as
\[=\frac{2secθtanθ+tan^2θ+(sec^2θ−1)}{sec^2θ+2secθtanθ+(tan^2θ+1)}\]
By \(sec^2x=tan^2x+1\), which can be derived from \(sin^2x+cos^2x=1\)
\[=\frac{2secθtanθ+tan^2θ+tan^2x}{sec^2θ+2secθtanθ+sec^2θ}\]

Answer 26

Ughhh
Another typo on the last line. It should be
\[=\frac{2secθtanθ+tan^2θ+tan^2\theta}{sec^2θ+2secθtanθ+sec^2θ}\]

Answer 27

lol its ok :)