Question

Can some one please help with this?: establish the identity: 1-cos/sin^2 = sec/1+sec

Answer 1

Should I assume such equation? \(\large{1-\frac{cos\theta}{sin^2\theta} = \frac{sec\theta}{1} + sec\theta}\)

Answer 2

Sorry it's 1-cos divided by sin^2 equals sec divided by 1+sec

Answer 3

just clear it\[\frac{1-\cos \theta}{
\sin^2 \theta}=\frac{\sec \theta}{1+\sec\theta}\]

Answer 4

@mukushla Ninja'd.

Answer 5

:)

Answer 6

id probably say multiply num and denum of LHS by \(1+\cos \theta\)

Answer 7

Came up with 1-cos^2 dived by sin^2+sin^2•cos but it doesn't look right

Answer 8

so u came up with\[\frac{1-\cos^2 \theta}{\sin^2 \theta(1+\cos\theta)}\]note that \(1-\cos^2 \theta=\sin^2 \theta\)

Answer 9

Yes

Answer 10

Not sure what to do next

Answer 11

well\[\frac{1-\cos^2 \theta}{\sin^2 \theta(1+\cos\theta)}=\frac{\sin^2 \theta}{\sin^2 \theta(1+\cos\theta)}=\frac{1}{1+\cos \theta}\]divide num and denum by \(\cos \theta\) and seee what happens

Answer 12

I'm probably wrong but came up with 1 divided by (1+cos)

Answer 13

thats right :)\[\frac{1}{1+\cos\theta}=\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta}+1}=\frac{\sec\theta}{1+\sec\theta}\]