(1 + cot + tan )(sin - cos ) whole upon sin^3 - cosec^3

[sambarbarian]

(1 + cot + tan )(sin - cos ) whole upon sin^3 - cosec^3 = sin^2cos^2

prove this . ( assume theta is front of all functions ) , i don't know where to start :/

 

[HallsofIvy]

I spent quite a long time working on that and then realized I had miscopied it!
And I wonder if you haven't miscopied it! Because what you have written cannot be proven- it is not true. If you take \theta= \pi/4 radians or 45 degrees, then sin(\theta)= cos(\theta)= \sqrt{2}/2 so that sin(\theta)- cos(\theta)= 0 and the numerator on the left is 0. We need to check that sin^3(\theta)- cosec^3(\theta)= 2^{3/2}/8- 8/2^{3/2} is not 0. Since it is not the left side is 0. But the right side is sin^2(\theta)cos^2(\theta)= \frac{1}{2}\frac{1}{2}= \frac{1}{4}, NOT 0.

Perhaps you miscopied? While I didn't finish, if the denominator were sin^3- cos^3 rather than sin^3- cosec^3, my counter-example would not work.

 

[sambarbarian]

no , i rechecked the question , it is correct and can be solved according to the publisher ( i called customer care )

 

[Mentallic]

no , i rechecked the question , it is correct and can be solved according to the publisher ( i called customer care )

What HallsofIvy is saying is that it can't be proven because it is not true. Whether some so-called customer care said so or not, it doesn't change the fact that mathematically they're not equivalent.

Just a couple of questions, is this what the problem says?

Prove that
\frac{(1+\cot \theta+\tan \theta)(\sin\theta - \cos\theta)}{\sin^3\theta-\csc^3\theta}=\sin^2\theta\cos^2\theta

If it is, then it can't be proven because there exist values of \theta such that the left-hand side does not equal to the right-hand side.

However, if the problem asked to solve rather than to prove, then we have a different question entirely. When the problem asks to solve for \theta what it's essentially asking is to find all the value(s) of \theta such that the left-hand side does equal the right-hand side. For example, \theta=0 satisfies the equality.

 

[sambarbarian]

yes , that is correct , but if it can't be solved , I am sorry, thank you for your time .

 

[Mentallic]

but if it can't be solved

It can be solved, but not proven to be an identity

You can solve x^2-2x+1=0 but you cannot prove that x^2-2x+1=0 because that implies that for every value of x, x^2-2x+1=0 which is untrue.

You can prove that \sin^2\theta+\cos^2\theta=1 which means if you solve for \theta you'll find that \theta can be any value to satisfy the condition that \sin^2\theta+\cos^2\theta=1

 

[HallsofIvy]

"Prove this" means "prove that this equation is correct for all x". That is not possible because it is not true as I said above. "Solve this equation" means find those particular values of x for which the equation is true. Is that what your problem asks you to do?