(1+sinx+cosx)/(1+sinx-cosx)=cotx/2?

Answer 1

1 + sinX + cosX (1 + sinX + cosX)²
---------------- = -----------------------------------
1 + sinX - cosX (1 + sinX - cosX)(1 + sinX + cosX)²

1 + 2sinX + 2cosX + 2sinXcosX + (sinX)² + (cosX)
= -----------------------------------------------------
1 + 2sinX + (sinX)² - (cosX)²

Now, notice almost everything has a coefficient of 2. Substituting 1
for (sinX)² + (cosX)² in the numerator and (sinX)² for 1 - (cosX)²
in the denominator gives

2 + 2sinX + 2cosX + 2sinXcosX
= -------------------------------
2sinX + 2(sinX)²

Factor by grouping in the numerator, and factor out sinX in the
denominator

(1 + sinX)(1 + cosX)
= ----------------------
(1 + sinX)sinX

Dividing numerator and denominator by 1+sinX gives

1 + cosX
= -----------
sinX

Answer 2

1. ( secx/sinx) - (sinx/cosx) = ( 1 / sin x cos x) - ( sinx / cos x) = ( 1 - sin^2 x)/ (sin x cos x) = cos^2 x / (sin x cos x) = cos x / sin x = cot x ans. 2. (1/1-cosx) + (1/1+cosx) = (1 + cosx + 1 - cosx ) / ( 1-cosx) ( 1+cosx) = 2 / ( 1 - cos^2 x) = 2 / sin^2 x = 2csc^2x

Answer 3

(1 + sinx + cosx) / (1 +sinx -- cosx)
= {sinx + (1 + cosx)} / {sinx + (1 -- cosx)}
= {2sinx/2cosx/2 + 2cos^2x/2} / {2sinx/2cosx/2 + 2sin^2x/2}
= 2cosx/2(sinx/2 + cosx/2) / 2sinx/2(cosx/2 + sinx/2)
= 2cosx/2 / 2sinx/2
= cotx/2

Answer 4

(1+ sin x + cos x)/(1+ sin x - cos x)

as RHS is in x/2 we reduce to x/2

sin x = 2 sin x/2 cos x/2
1+ cos x = 2 cos ^2 x/2

so 1+ cos x + sin x = 2 cox x/2(sin x/2+ cos x/2)

1- cos x = 2 sin ^2 x/2
1- cos x + sin x = 2 sin ^2 x/2 + 2 sin x/2 cos x/2
= 2 sin x/2(sin x/2 + cos x/2)
by division we get the result

Answer 5

2sinx/2 cosx/2 + 2cos^2x/2
-------------------------------------
2sinx/2 cosx/2 + 2sin^2x/2

= cotx/2

Answer 6

i would think that it means -tan...but thats just me....i kinda forgot the whole calculus thing

Answer 7

not complete