Question

Trigonometric Identities anyone? Need help....

Answer 1

On what?

Answer 2

2sec^2 beta= 1/1-sin beta+ 1/1+sin beta

Answer 3

Is this the equation?
\[
2\sec^2(\beta)=\frac{1}{1-\sin(\beta)}+\frac{1}{1+\sin(\beta)}
\]

Answer 4

yup

Answer 5

I would combine the two fractions together first.

Answer 6

how?

Answer 7

Another hint:
\[
\frac{1}{1-\sin(\beta)}\underbrace{\frac{1+\sin(\beta)}{1+\sin(\beta)}}_{=1}=?
\]

Answer 8

Still dont get it..........................................................

Answer 9

Do you know that \((1-x)(1+x)=1-x^2\)?

Answer 10

So this is mixed algebra and trigonometry?

Answer 11

Yep!

Answer 12

The denominator becomes negative?

Answer 13

Negative denominator is not a problem. The denominator may not be negative too. Consider \(\sin(0)=0\) and \(1-\sin(0)\).

Answer 14

Its cos(0) ryt?

Answer 15

Where does that \(\cos(0)\) come from?

Answer 16

1-sin(0)=cos(0)

Answer 17

hello?

Answer 18

Do you know the identity \(\cos^2(x)+\sin^2(x)=1\)? We have to use that.

Answer 19

pythagorean identity?

Answer 20

\(2\sec^2(\beta)=\dfrac{2}{\cos^2(\beta)}\) and \(\cos^2(\beta)=1-\sin^2(\beta)\). Any ideas now?

Answer 21

Then what becomes of 2/cos^theta?

Answer 22

yo thomas