Question

How do you simplify it??

Answer 1

Please see a Proof of Ex. 15 in the Explanation.

Explanation:

Let us solve Ex. 15.

We will use these Identities :

`(1+cosalpha)=2cos^2(alpha/2), (1-cosalpha)=2sin^2(alpha/2)`.

`2sin(alpha/2)cos(alpha/2)=sinalpha`.

Method 1 :

`sqrt{(1-cosalpha)/(1+cosalpha)}+sqrt{(1+cosalpha)/(1-cosalpha)}`,

`={(1-cosalpha)+(1+cosalpha)}/sqrt{(1+cosalpha)(1-cosalpha)}`,

`=2/sqrt(1-cos^2alpha)`,

`=2/|sinalpha|`,

`rArr sqrt{(1-cosalpha)/(1+cosalpha)}+sqrt{(1+cosalpha)/(1-cosalpha)}=2|cscalpha|`.

Method 2 :

`sqrt{(1-cosalpha)/(1+cosalpha)}+sqrt{(1+cosalpha)/(1-cosalpha)}`,

`=sqrt{(2sin^2(alpha/2))/(2cos^2(alpha/2))}+sqrt{(2cos^2(alpha/2))/(2sin^2(alpha/2)}`,

`=|sin(alpha/2)/cos(alpha/2)|+|cos(alpha/2)/sin(alpha/2)|`,

`={|sin(alpha/2)|^2+|cos(alpha/2)|^2}/(sin(alpha/2)*cos(alpha/2))`,

`=1/|(sin(alpha/2)*cos(alpha/2))|xx2/2`,

`=2/(2|sin(alpha/2)*cos(alpha/2)|)`,

`=2/|sin(2*alpha/2)|`,

`=2/|sinalpha|`, as before!

Ex. 16 can be proved in the similar fashion.

Since, the L.H.S. is non-negative, so has to be the

R.H.S.

Answer 2

# sqrt{ {1-cos alpha}/{ 1 + cos alpha} } + sqrt{ {1 +cos alpha}/{ 1 - cos alpha} } = | 2 csc alpha | #

Explanation:

I'm geeky enough to recognize these forms as associated with the tangent half angle formulas.

The first stem from the cosine double angle formula:

`cos 2x = cos ^2 x - sin ^2 x = 2 cos ^2 x -1 = 1 - sin ^2 x`

These provide a nice way to change 1 plus or minus cosine to twice the square of the trig function of a half angle. It's easier in symbols:

`2 cos ^2 x = 1 + cos 2 x`

`2 sin ^2 x = 1 - cos 2 x`

We get

`sqrt{ {1-cos alpha}/{ 1 + cos alpha} } = \sqrt{ { 2 sin ^2 ( alpha/2 ) } / { 2 cos ^2 (alpha/2)} } = \sqrt{ tan^2 (alpha/2) } = |tan(alpha/2)|`

I always get concerned when the absolute value appears. But w're after the sum of principal values, positive numbers, so let's go with it.

`sqrt{ {1-cos alpha}/{ 1 + cos alpha} } + sqrt{ {1 +cos alpha}/{ 1 - cos alpha} }`

`= |tan(alpha/2)| + |1/ tan(alpha/2)|`

The tangent and its reciprocal have the same sign so we can bring them together like this:

`=| tan(alpha/2) +1/ tan(alpha/2)|`

`=| { 1 + tan^2(alpha/2) }/ tan(alpha/2)|`

`=| { sec^2(alpha/2) }/ tan(alpha/2)|`

`= | 1/ cos^2(alpha/2) cdot cos(alpha/2)/sin(alpha/2) |`

`= | 1/ { cos(alpha/2) sin(alpha/2) } |`

`= | 2/ sin alpha |`

`= | 2 csc alpha |`

Wow, that simplified.

`sqrt{ {1-cos alpha}/{ 1 + cos alpha} } + sqrt{ {1 +cos alpha}/{ 1 - cos alpha} } = | 2 csc alpha |`

Let's check it. Let's see what the Socratic grapher does with it. We'll plot

# sqrt{ {1-cos alpha}/{ 1 + cos alpha} } + sqrt{ {1 +cos alpha}/{ 1 - cos alpha} } - | 2 csc alpha | #

graph{sqrt((1-cos x)/(1 + cos x)) + sqrt((1 +cos x)/( 1 - cos x ) ) - | 2 csc x| [-0.00782, 0.007924, -0.0003786, 0.0004086]}

That's zero, though it does get hinky around the singularities at `alpha=k pi`.

Gotta go, can't do the other one right now.