trouble proving trig identity, solving trig equation

[alphaomega]

I received a packet of about 20 problems two weeks ago to solve and I've had trouble on two of them. I've tried a million different ways by using the trig identities but I'm still stuck:

1) Prove that:

2csc(x) = 1 + sec(x) / tan(x) + tan(x) / 1 + sec(x)

On the following one, I'm extremely confused, because there are 6 answers.

2) Solve (has six answers)

csc^2(x) - 2csc(x) = 2 - 4 sin(x)

 

[galactus]

Re: Trouble with trig

Solve (has six answers)

csc^2(x) -2csc(x) = 2 - 4 sin(x)

\(\displaystyle \L\\csc^{2}(x)-2csc(x)=2-4sin(x)\)

\(\displaystyle \L\\\frac{1}{sin^{2}(x)}-\frac{2}{sin(x)}=2(1-2sin(x))\)

\(\displaystyle \L\\\frac{sin(x)-2sin^{2}(x)}{sin^{3}(x)}=2(1-2sin(x))\)

\(\displaystyle \L\\\frac{(1-2sin(x))}{sin^{2}(x)}=2(1-2sin(x))\)

Can you finsih up?.

 

[alphaomega]

thanks alot that helped me a bunch , can anyone lend me some help for the first problem

 

[soroban]

Hello, alphaomega!

1) Prove that: \(\displaystyle \L\:2\,\csc(x) \;=\;\frac{1\,+\,\sec(x)}{\tan(x)}\,+\,\frac{\tan(x)}{1\,+\,\sec(x)}\)

On the right side, multiply the second fraction by \(\displaystyle \frac{\sec(x) - 1}{\sec(x)-1}\)

We have: \(\displaystyle \L\:\frac{\tan(x)}{\sec(x)\,+\,1}\,\cdot\,\frac{\sec(x)\,-\,1}{\sec(x)\,-\,1} \:=\:\frac{\tan(x)[\sec(x)\,-\,1]}{\sec^2(x)\,-\,1}\)

. . . . . . . . \(\displaystyle \L=\:\frac{\tan(x)[\sec(x)\,-\,1]}{\tan^2(x)} \:=\:\frac{\sec(x)\,-\,1}{\tan(x)}\)


The right side becomes: \(\displaystyle \L\:\frac{1\,+\,\sec(x)}{\tan(x)}\,+\,\frac{\sec(x)\,-\,1}{\tan(x)} \;=\;\frac{2\sec(x)}{\tan(x)}\)

. . . . . . . \(\displaystyle \L=\:\frac{2\left(\frac{1}{\cos(x)}\right)}{\frac{\sin(x)}{\cos(x)}} \;=\;\frac{2}{\sin(x)} \;=\;2\,\csc(x)\)