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!
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!)
3 answers
(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
(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
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
( 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