sin theta/csc theta +cos theta/sec theta =1?

Answer 1

sinө/(1/sinө) + cosө(1/cosө) =
sin²ө + cos²ө = 1 (obvious)

Answer 2

Hey there!

Here's the answer.

sin(θ)/csc(θ)+cos(θ)/sec(θ)=1 --> Write the problem.
sin(θ)/(1/sin(θ))+cos(θ)/(1/cos(θ)) --> Use the identities sec(θ)=1/cos(θ) and csc(θ)=1/sin(θ).
sin^2(θ)+cos^2(θ)=1 --> Divide and simplify the above equation.
1=1, true Use the Pythagorean identity.

So that was the proof on why sin(θ)/csc(θ)+cos(θ)/sec(θ)=1.

Here are some of the identities.

sin^2(θ)+cos^2(θ)=1
1+tan^2(θ)=sec^2(θ).
cot^2(θ)+1=csc^2(θ).

Those where the Pythagorean identities.

Hope it helps!

ADDITION: If you're solving for θ, the answer would be all real numbers, in the interval -2pi<θ<2pi. This interval is a sample, but to me the answer is all real numbers.

Answer 3

Prove the identity.

sinθ/cscθ + cosθ/secθ = 1

Let's start with the left hand side.

Left Hand Side = sinθ/cscθ + cosθ/secθ

= (sinθ)(sinθ) + (cosθ)(cosθ)

= sin²θ + cos²θ = 1 = Right Hand Side.

Answer 4

sin θ/csc θ + cos θ/sec θ = 1 -->Prove??

Start with the more complicated side (left):

sin θ/csc θ + cos θ/sec θ
= sin θ/(1/sin θ) + cos θ/(1/cos θ)
= sin θ*(sin θ/1) + cos θ*(cos θ/1)
= sin² θ + cos² θ
= 1



Note - above I used known basic identities:
csc θ = 1/sin θ
sec θ = 1/cos θ
sin² θ + cos² θ = 1