Posted by **AwesomeGuy** on Thursday, February 14, 2013 at 12:34am.

Verify the identities.

1.) SIN[(π/2)-X]/COS[(π/2)-X]=COT X

2.) SEC(-X)/CSC(-X)= -TAN X

3.) (1 + SIN Y)[1 + SIN(-Y)]= COS²Y

4.) 1 + CSC(-θ)/COS(-θ) + COT(-θ)= SEC θ

(Note: Just relax through verifying/solving these nice fun looking math problems! It's healthy for your brain!)

- Trigonometry -
**AwesomeGuy**, Thursday, February 14, 2013 at 3:47am
1.) 1/TAN[(π/2)-X]=COT X BINGO! SOLVED!

2.) SEC X/-CSC X

1/COS X ÷ -1/COS X

1/COS X * -SIN X/1

-TAN X YES BINGO! WOW!

- Trigonometry -
**Steve**, Thursday, February 14, 2013 at 4:42am
(1+sin(y))(1+sin(-y)

(1+sin(y))(1-sin(y))

(1-sin^2(y))

cos^2(y)

I think the last one has a typo or needs some parentheses. If θ=pi/4,

1 + (-√2)/(1/√2) + (-1) = 1 - 2 - 1 = -2

but sec(pi/4) = √2

- Trigonometry -
**Reiny**, Thursday, February 14, 2013 at 7:43am
I think the last one should be

( 1 + csc(-Ø) / ( cos(-Ø) + cot(-Ø) ) = secØ

First of all , csc(-x) = -cscx and cot(-x) = -cotx , but cos(-x) = cosx

LS = (1 - cscx)/( cosx - cotx)

= (1 - 1/sinx) / (cosx - cosx/sinx)

= [ (sinx - 1)/sinx ] / [ (sinxcosx - cosx)/sinx ]

= (sinx - 1) / (sinxcosx - cosx)

= (sinx - 1) / (cosx(sinx - 1) )

= 1/cosx

= secx

= RS

## Answer this Question

## Related Questions

- Trigonometry - Verify the identities. 1.) √1-COSθ/1+COSθ= 1+SIN...
- Math - Calculus - The identity below is significant because it relates 3 ...
- Trigonometry - 1.Solve tan^2x + tan x – 1 = 0 for the principal value(s) to two ...
- Precalc/Trig - Sorry there are quite a few problems, but I just need to know if ...
- Math Question - Given tan θ = -8/5 and sin θ < 0, find sin θ, ...
- Trigonometry - Find the remaining trigonometric ratios of θ if sin(θ) ...
- pre-calc help !!!!!!! - 1) cscθsinθ-sin^2θ 2)sinθcscθ/...
- Trig - Given that csc θ = -4 and tan θ > 0, find the exact value of...
- trigonometry hlp me - 6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y...
- Math Question - Find the values of the six trigonometric functions of an angle ...

More Related Questions