Trigonometric identities?

Question

How can I demostrate that:

sin + sin * tan + cos * cot + cos = csc * sec

I just got this:

sin + sin * sin/cos + cos * cos/sin + cos

But i'm stuck after that.

I have tried to solve this but I just cant' do it. Already tried to change everything to sin, cos, but it didn't work. Any suggestions?

You are right, I was missing a + sign, thanks a lot for your answer.

sin + sin * tan + cos * cot + cos = csc + sec

Answer 1

sin φ + sin φ tan φ + cos φ cot φ + cos φ = csc φ sec φ

sin φ + sin φ (sin φ / cos φ) + cos φ (cos φ / sin φ) + cos φ = csc φ sec φ
sin φ + sin²φ / cos φ + cos²φ / sin φ + cos φ = csc φ sec φ

getting the least common denominator which is sin φ cos φ

then

(sin²φ cos φ + sin²φ sin φ + cos²φ cos φ + sin φ cos²φ) / sin φ cos φ = csc φ sec φ

combining similar terms

(sin²φ cos φ + cos²φ cos φ) + (sin²φ sin φ + cos²φ sin φ) / sin φ cos φ = csc φ sec φ

then

[cos φ(sin²φ + cos²φ) + sin φ(sin²φ + cos²φ)] / sin φ cos φ = csc φ sec φ

sin²φ + cos²φ = 1

(cos φ + sin φ) / sin φ cos φ = csc φ sec φ

1/sin φ + 1/cos φ = csc φ sec φ

cscφ + sec φ = cscφ sec φ

i think you forgot to place a plus sign in between the csc φ and sec φ. i you do, then my solution is right. hope this will help.

Answer 2

I don't think that equation is correct. Try plotting it. It also has singularities unless the range is limited. I think magina is right: the right side should be csc + sec; Expanding tan, cot, sec, and csc you get

sin + sin^2/cos +cos^2/sin + cos = 1/cos + 1/sin

Then use sin^2 = 1-cos^2 and cos^2 = 1-sin^2 to get


sin + (1-cos^2)/cos +(1-sin^2)/sin + cos = 1/cos + 1/sin

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


1/cos + 1/sin = 1/cos + 1/sin QED

Answer 3

sin + sin * tan + cos * cot + cos = csc * sec

sin + sin * sin/cos + cos * cot =

sin + sin^2 / cos + cos^2/ sin + cos =

Multiply by the common denominator which is cos * sin

sin^2cos + sin^3 + cos^3 + cos / cos * sin

That's the farthest I can get...I don't know what to do afterwards