Use trigonometric identities to show:

Part a is not an identity.

If you try to establish:  sin(t) / ( 1 + cos(t) )   =   csc(t)

--->     sin(x) / ( 1 + cos(t) )  ·  [ ( 1 - cos(t) ) / ( 1 - cos(t) )             

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ ( 1 + cos(t) ) · ( 1 - cos(t) ) ]

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ 1 - cos²(t) ]                       since  1 - cos²(t) =  sin²(t) --->

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ sin²(t)  ]       

--->     ( 1 - cos(t) ) / sin(t)

--->     1 / sin(t) - cos(t) / sin(t)

--->     csc(t) - cot(t)                                        not just csc(t) 

Part b:      ( sec²(t) - 1 ) / sec²(t)  =  sin²(t)

--->     ( sec²(t) - 1 ) / sec²(t)

--->     sec²(t) / sec²(t)  -  1 / sec²(t)

--->     1 - cos²(t)

--->     sin²(t)

Part c:     Write 1 / [ 1 + sin(x) ]  without fractions:

--->     1 / [ 1 + sin(x) ]  ·  [ ( 1 - sin(t) ) / ( 1 - sin(t) ) ]

--->     [ 1 · ( 1 - sin(t) ) ] / [ ( 1 + sin(t) ) · ( 1 - sin(t) ) ]

--->     [ 1 - sin(t) ] / [ 1 - sin²(t) ]

--->     [ 1 - sin(t) ] / [ cos²(t) ]

--->     1 / cos²(t)  -  sin(t) / cos²(t)

--->     sec²(t)  -  sin(t) / cos(t) · 1 / cos(t)

--->     sec²(t)  - tan(t) ·sec(t)

Hello Calpal!

Use trigonometric identities to show:

a) Sin(t) / (1+cos(t)) = csc(t)

b) (sec2(t)-1) / sec2(t) = sin2(t)

c) Write an equivalent form of 1 / (1+sin(x)) without using fractions

First off a)

a)

sin(t) / (1+cos(t)) = csc(t)

sin(t) / (1+cos(t)) = 1 / sin(t)

sin²(t) = 1 + cos(t)

1- cos²(t) = 1 + cos(t)

cos²(t) = - cos(t)

t ∈ {± periodically π/2; π; 3π/2}

Greeting asinus :- )  !

c) Write an equivalent form of 1 / (1+sin(x)) without using fractions.

Multiply top/bottom by (1 - sin x) and we have

[1 - sinx] / [ 1 - sin^2x]  =

[1 - sin x ] / cos^2x =

1/cos^2x - sinx/ cos^2x =

sec^2x - tanx* secx =

secx [ secx - tan x]