Trig Identities Help!

[Xelaxja]

I have been trying to do these questions for a while. I just am not getting it. Here is what I have tried..

Prove the following Identities.
1. (secA-tanA)(cosecA+1)= cotA

I started by putting everything in cosine and sine.

1/cosA -(SinA/CosA)(1/SinA+1)

From there I go

(1-SinA)/CosA(1/SinA+1)

I just get stuck....

2.cosAcosexA+SinAsecA= cosecA secA

Once again everything into cosine and sine.

cosA(1/sinA)+Sin(1/cosX)

Then I simplify it further to...

cos/Sin + Sin/Cos

Can you believe I figured it out while typing it here? Well at least I think....confirm please.

Previously I converted to Tan then performed the adittion.

Now I did it as sine and cosine.

Cos/Sin+ Sin/Cos

Cos^2+Sin^2
------------------------
CosSin

Resulting in....

1/CosSin

Then...

1/Cos + 1/Sin

Therefore....

cosecASecA

Right?


3. secAcosecA-cotA= tanA

I first do the usual conversion into sine and cosine.

(1/cosA) * (1/sinA) - (cosA/sinA)

Multiplication...

1/(cosAsinA) - ( cosA/sinA)

Addition.

1-cosA/(cosASinA)

and....

I'm stuck....Well here it is...please help soon. I just can't get them out. I tried for a loooong time. >.>

 

[stapel]

Prove the following Identities.
1. (secA-tanA)(cosecA+1)= cotA

I started by putting everything in cosine and sine.

Good idea! But I'm afraid I don't follow your steps...? Instead, for the left-hand side, try:

. . . . .[1/cos(A) - sin(A)/cos(A)] [1/sin(A) + 1]

. . . . .[(1 - sin(A))/cos(A)][(1 + sin(A))/sin(A)]

. . . . .[(1 - sin(A))(1 + sin(A))] / [cos(A)sin(A)]

See where that leads.

2.cosAcosexA+SinAsecA= cosecA secA

Once again everything into cosine and sine.

cosA(1/sinA)+Sin(1/cosX)

Cos/Sin+ Sin/Cos

Cos^2+Sin^2
------------------------
CosSin

1/CosSin

Then...

1/Cos + 1/Sin

Um... How did the product turn into a sum...? :shock:

Note: Without the arguments, the functions are ill-defined, and your solution will probably be counted as incorrect. The reasoning behind this will become clear when you start working with identities with "2a" or "a/2", etc, as the arguments of some (and not all) of the functional terms.

Try this on the left-hand side:

. . . . .cos(A)/sin(A) + sin(A)/cos(A)

. . . . .cos[sup:yeehuvnn]2[/sup:yeehuvnn](A) / [sin(A)cos(A)] + sin[sup:yeehuvnn]2[/sup:yeehuvnn](A) / [sin(A)cos(A)]

. . . . .[cos[sup:yeehuvnn]2[/sup:yeehuvnn](A) + sin[sup:yeehuvnn]2[/sup:yeehuvnn](A)] / [sin(A)cos(A)]

Then apply the Pythagorean Identity to the numerator, and work "backwards" from the resulting expression to the expression on the right-hand side of the original equation.

3. secAcosecA-cotA= tanA

I first do the usual conversion into sine and cosine.

(1/cosA) * (1/sinA) - (cosA/sinA)

1/(cosAsinA) - ( cosA/sinA)

1-cosA/(cosASinA)

and....I'm stuck....

A common denominator is often (though not always) quite helpful. Working on the left-hand side:

. . . . .1/[sin(A)cos(A)] - cos(A)/sin(A)

. . . . .1/[sin(A)cos(A)] - cos[sup:yeehuvnn]2[/sup:yeehuvnn](A)/[sin(A)cos(A)]

. . . . .[1 - cos[sup:yeehuvnn]2[/sup:yeehuvnn](A)] / [sin(A)cos(A)]

Now apply the Pythagorean Identity, simplify, and convert the result to the expression on the right-hand side of the original equation.

By the way, thank you for showing your work and reasoning so clearly!

Eliz.