Trig-Establishing/Verifying Identities

[Variance]

I'd really appreciate it if somebody could help me these these. I'm just stumped.

1/(csc(x)-cot(x)) = csc(x)+cot(x)

and

1+sin^2(x)=sec^2(x)+sin^2(x)-sin^2(x)*sec^2(x)

 

[stapel]

For the first one, work from the left-hand side. Convert everything to sines and cosines, combine the two terms in the denominator, and (now that you're dividing by a fraction), flip and simplify. Simplify the denominator by multiplying by the conjugate. Apply the Pythagorean Identity to convert the denominator from cosines to sines. Then factor the numerator, cancelling with one of the factors in the denominator. Simplify again, and you should get the right-hand side.

Eliz.

 

[Variance]

D'oh! The thought of using a conjugate completely slipped my mind. Thanks for the help; now I've solved that one. Now if I can just get the second one knocked out...

 

[soroban]

Hello, Variance!

\(\displaystyle 1\,+\,\sin^2(x)\:=\:\sec^2(x)\,+\,\sin^2(x)\,-\,\sin^2(x)\cdot\sec^2(x)\)

The right side is: \(\displaystyle \,\sec^2(x)\,-\,\sin^2(x)\cdot\sec^2(x)\,+\,\sin^2(x)\)

Factor: \(\displaystyle \,\sec^2(x)[\underbrace{1\,-\,\sin^2(x)}]\,+\,\sin^2x\)

Then: \(\displaystyle \;\;\;\;\underbrace{\sec^2(x)\cdot\cos^2x}\,+\,\sin^2(x)\)

Finally: \(\displaystyle \;\;\;\;\;\;\;\;\;1\,+\,\sin^2(x)\)

 

[Mrspi]

D'oh! The thought of using a conjugate completely slipped my mind. Thanks for the help; now I've solved that one. Now if I can just get the second one knocked out...

You have
1 + sin<SUP>2</SUP> x = sec<SUP>2</SUP> x + sin<SUP>2</SUP> x - sin<SUP>2</SUP> x sec<SUP>2</SUP> x

Work with the right-hand side (it is often easier to simplify a complicated expression than it is to "complicate" a simpler expression). Let's rearrange the terms:

sec<SUP>2</SUP> x - sin<SUP>2</SUP> x sec<SUP>2</SUP> x + sin<SUP>2</SUP> x

Factor sec<SUP>2</SUP> x out of the first two terms:

sec<SUP>2</SUP> x (1 - sin<SUP>2</SUP> x) + sin<SUP>2</SUP> x

Now....can you take it from here?

 

[Variance]

I got it. Thanks for the help, everyone! Once it was pointed out that I could factor part of it, it all became clear.