Question

Find the value of:


\(\huge\color{blue}{(\frac{sec(\theta) + tan(\theta) - 1}{tan(\theta) - sec(\theta) + 1})^2}\).

Answer 1

@Directrix @hartnn

Answer 2

@jim_thompson5910 @campbell_st @dumbcow

Answer 3

Expand then use some trigo identities to simplify it. The answer is sec(theta)+tan(theta).

Answer 4

Certainly I don't want answer, I have answer choices with me, I just want the solution..
Simply tell me what formula to use here..

Answer 5

oops, sorry, should be (sec(theta)+tan(theta))^2

Answer 6

Never mind, don't concentrate on answer...

Answer 7

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

Answer 8

sec(theta)=1/cos(theta) and tan(theta)=sin(theta)/cos(theta), first step is trying to rewrite the given expression in terms of sin(theta) and cos(theta)

Answer 9

\[=\frac{1+\sin\theta}{1-\sin\theta}\]

Answer 10

I am sure I might have messed up somewhere in the tex!!

Answer 11

Yep you are right electrokid... Thank you so much..

My mind was not working on it earlier.

Answer 12

yay!

Answer 13

Let me check it now carefully, thanks to all for coming to atleast see this post..

Answer 14

when a smrty with score 99 posts a question, we expect some fun :D

Answer 15

Was this not fun??