Trig Indentity Problem

[thisisinvisible]

If tan x + cot x = 4, what is the value of:

sin[sup:1fu6f5w9]2[/sup:1fu6f5w9]x + cos[sup:1fu6f5w9]2[/sup:1fu6f5w9]x + tan[sup:1fu6f5w9]2[/sup:1fu6f5w9]x + cot[sup:1fu6f5w9]2[/sup:1fu6f5w9]x + sec[sup:1fu6f5w9]2[/sup:1fu6f5w9]x + csc[sup:1fu6f5w9]2[/sup:1fu6f5w9]x?


I know the answer is supposed to be 31, but I can't seem to get that answer. I would appreciate it if someone showed the steps to get 31.

 

[tutor_joel]

That's a neat little identity,

The trick is to use the information that they give you and manipulate it to fit into the long statement. To get started, take the tan x + cot x = 4 equation and multiply through by tan x and cot x separately to yield:

\(\displaystyle tan^2x + 1 = 4tan x\)

and

\(\displaystyle 1+ cot^2x = 4cot x\)

substitute these values into the big equation and see if that helps

 

[garf]

If you use the fundamental identities (sen[sup:1ai6les2]2[/sup:1ai6les2](x)+cos[sup:1ai6les2]2[/sup:1ai6les2](x) =1, 1+tan[sup:1ai6les2]2[/sup:1ai6les2](x)=sec[sup:1ai6les2]2[/sup:1ai6les2](x), etc) and the hint that Tutor_joel gave you,you can find the answer (31). It is just to play with the trig. identities.

 

[thisisinvisible]

That's a neat little identity,

The trick is to use the information that they give you and manipulate it to fit into the long statement. To get started, take the tan x + cot x = 4 equation and multiply through by tan x and cot x separately to yield:

\(\displaystyle tan^2x + 1 = 4tan x\)

and

\(\displaystyle 1+ cot^2x = 4cot x\)

substitute these values into the big equation and see if that helps

That seemed to help, though I'm still a bit stuck. Here's what I did:

sin[sup:267s72lg]2[/sup:267s72lg]x + cos[sup:267s72lg]2[/sup:267s72lg]x + tan[sup:267s72lg]2[/sup:267s72lg]x + cot[sup:267s72lg]2[/sup:267s72lg]x + sec[sup:267s72lg]2[/sup:267s72lg]x + csc[sup:267s72lg]2[/sup:267s72lg]x (Original Problem)

1 + tan[sup:267s72lg]2[/sup:267s72lg]x +cot[sup:267s72lg]2[/sup:267s72lg]x +sec[sup:267s72lg]2[/sup:267s72lg]x + csc[sup:267s72lg]2[/sup:267s72lg]x

4tanx + cot[sup:267s72lg]2[/sup:267s72lg]x + sec[sup:267s72lg]2[/sup:267s72lg]x + csc[sup:267s72lg]2[/sup:267s72lg]x

4tanx + cot[sup:267s72lg]2[/sup:267s72lg]x + (1+tan[sup:267s72lg]2[/sup:267s72lg]x) + (1 + cot[sup:267s72lg]2[/sup:267s72lg]x)

4tanx + cot[sup:267s72lg]2[/sup:267s72lg]x + 4tanx + 4cotx

8tanx + 4cotx + (4cotx - 1)

8tanx + 8cotx - 1



And that's where I left off. I made a bunch of substitutions using your help and Pythagorean identities. Did I go wrong somewhere or did I just not go far enough?

 

[tutor_joel]

That's the answer. Factor out the 8 and you'll have 8*4 - 1 = 31

 

[soroban]

Hello, thisisinvisible!

Another approach . . .

\(\displaystyle \text{If }\tan x + \cot x \:=\: 4,\: \text{evaluate: }\: \sin^2\!x + \cos^2\!x + \tan^2\!x + \cot^2\!x + \sec^2\!x + \csc^2\!x\)

\(\displaystyle \text{Note that: }\:\tan x + \cot x \:=\:4\)

. . \(\displaystyle \text{Then: }\\tan x + \cot x)^2 \:=\:16 \quad\Rightarrow\quad \tan^2\!x + 2 + \cot^2\!x \:=\:16\)

. . \(\displaystyle \text{Hence: }\:\tan^2\!x + \cot^2\!x \:=\:14\)


\(\displaystyle \text{We have: }\;\bigg[\sin^2\!x + \cos^2\!x\bigg] + \bigg[\tan^2\!x + \cot^2\!x\bigg] + \bigg[(\tan^2\!x+1) + (\cot^2\!x+1)\bigg]\)

. . . . . . . . \(\displaystyle =\; \underbrace{\bigg[\sin^2\!x + \cos^2\!x\bigg]}_{\text{This is 1}} + \underbrace{\bigg[\tan^2\!x + \cot^2\!x\bigg]}_{\text{This is 14}} + \underbrace{\bigg[\tan^2\!x + \cot^2\!x\bigg]}_{\text{This is 14}} +\; 2\)

. . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle =\quad31\)