csc(90º+θ) = secθ?

Question

how do you figure it out?

Answer 1

Ok...let's see,

csc(theta) = 1/sin(theta)

sin (90 + theta) = sin(90) cos(theta) + sin(theta)cos(90)
= cos(theta)

so,

1/sin(90+theta) = 1/cos(theta) = sec(theta)

Good luck!

Answer 2

From the left :

= csc(90º+θ)
= 1 / sin( 90º + θ) ( from definition of csc)
= 1 / cos (θ) ( sin( 90º + θ) = cos(θ) )
= secθ (definition for secθ)

and that equals the right side. That's it !

Answer 3

csc(90º + θ) = sec θ

To solve, use the reciprocal identity for csc
1/sin(90º + θ) = sec θ

Use the addition formula sin (A + B) = sin A cos B + cos A sin B
1/(sin 90º cos θ + cos 90º sin θ) = sec θ

Use the following identities: sin 90º = 1 and cos 90º = 0.
1/(1 · cos θ + 0 · sin θ) = sec θ

Thus,
1/cos θ = sec θ

Therefore,
sec θ = sec θ. QED.

^_^

Answer 4

csc θ = 1/sin θ i.e. csc(90º+θ) = 1/sin(90º+θ)
sec θ = 1/cos θ
So, we have,
1/sin(90º+θ) = 1/cos θ
or, cos θ = sin (90º+θ)

I hope this helps

Answer 5

csc=1/sin
sin (a+b)=sin a *cos b +cos a *sin b
sin (90+t)=sin 90 * cos t + cos 90 * sin t
sin 90=1
cos 90=0
sin (90+t)=cos t
csc(90+t)=1/sin(90+t)=1/cos t=sec t

Answer 6

csc(90 + A) = sec(A)
1/(sin(90 + A)) = 1/cos(A)

you can also write it as

sin(90 + A) = cos(A)

sin(90 + A) =
sin(90)cos(A) + cos(90)sin(A) =
1cos(A) + 0sinA =
cosA

so

sin(90 + A) = cos(A)

or in our case

csc(90 + A) = sec(A)

Answer 7

csc(90+θ)=secθ
1/(sin(90+θ))=secθ
1/(sin90cosθ+cos90sinθ)=secθ
1/(1*cosθ+0*sinθ)=secθ
1/cosθ=secθ
secθ=secθ

solved!!!