SOLUTION 4: Homework

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Pick a side to work on and then break things down and work on it one at a time. Remember to not bring anything over the equal sign because when you're proving something they should work out the same if they are equal.

Break things down so it's easier and work on simplifying the broken parts of the logarithmetric (merge of the words logarithm and trigonometric) identity.

1. Break the logarithmetric expression up to make things easier.
2. Now you have three different trigonometric identities on the left hand side.
3. Work out the trigonometric identities one by one.
4. The first expression in the denominator it has csc²θ - cot²θ - sin²θ. If you notice csc²θ - cot²θ is a trigonometric identity which equals to one.
   csc²θ - cot²θ = 1
5. Now in the denominator we have 1-sin²θ. That is also a trigonometric identity:
   1-sin²θ = cos²θ
6. To simplify things we have sin²θ/ cos²θ which is also the same thing as tan²θ.

Moving on to the next trigonometric identity to solve:

1. In the numerator we have sec²θ - tan²θ - cos ²θ
2. The trigonometric sec²θ - tan²θ = 1
3. So now, what we have in the numerator is 1 - cos²θ
4. That is also a trigonometric identity is 1 - cos²θ = sin²θ
5. Now we are left with sin²θ/cos²θ and that is the same thing as tan²θ

Now we can re-write the expression with the new simplified trig identity. Now, it is the log of base tan²θ to the power of tan²θ to the exponent of a trigonometric identity. Since the base is the same as the base of argument, that must mean that the whole log is equal to the exponent of argument.

Now pick one side to work on and try to transform that side into looking identical to the other side. The easiest side would be the left hand side because if you look carefully there's a lot more you can do to that side.

1. In the numerator we have csc²θ + cos(2θ). The trigonometric identity of cos(2θ) is equal to cos²θ - sin²θ.
2. Now what we have in the numerator is csc²θ + cos²θ - sin²θ.
3. In the denominator we have cot²θ + sin³θ and cot²θ is the exact same as cos²θ/sin²θ
4. Now to add the denominator both fractions have to have the same denominators. We multiply the bottom and top by sin²θ. Now what we have in the denominator is cos²θ-sin^5θ all over sin²θ.
5. The numerator is csc²θ + cos²θ - sin²θ. csc²θ is the same thing as 1/sin²θ.
   
6. To add ever thing together we have to get all the others denominators equal to sin²θ.
7. Now what we will have in the numerator is 1 + cos²θsin²θ - sin^4θ all over sin²θ.
8. What we have now is an improper fraction. So what we do is take the reciprocal of one fraction and multiply it to the other. The sin²θ reduce. What we have now is 1 + cos²θsin²θ - sin^4θ all over cos²θ-sin^5θ.
9. On the top cos²θ is the same thing as 1- sin²θ. In the numerator it'll be 1+ sin²θ(1-sin²θ) - sin^4θ.
10. When the numerator is expanded we have 1+ sin²θ - sin^4θ - sin^4θ, which is 1+ sin²θ - 2sin^4θ.
11. Q.E.D we have the exact same thing on the right hand side of the equation. The identity is now 1+ sin²θ - 2sin^4θ / cos²θ-sin^5θ.