Simplify.

[rishikachaks]

I have tried simplifying it but got nowhere near any of the 4 multiple-choice options. Thank you in advance.

 

[Deleted member 4993]

View attachment 27336

I have tried simplifying it but got nowhere near any of the 4 multiple-choice options. Thank you in advance.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[skeeter]

[MATH]\cot^2\left(\frac{\pi}{2}-x\right) = \tan^2{x}[/MATH]
[MATH]\sec^2(-x) = \sec^2{x} = \tan^2{x} +1[/MATH]
try again …

 

[nasi112]

It is fun to solve this problem, but if you are in a test, and you feel that you don't have enough time to think, just plug numbers in your calculator and see which one matches the equation.

 

[Steven G]

It is fun to solve this problem, but if you are in a test, and you feel that you don't have enough time to think, just plug numbers in your calculator and see which one matches the equation.

There is always time to think. That is what a test is for!

 

[lex]

It is fun to solve this problem, but if you are in a test, and you feel that you don't have enough time to think, just plug numbers in your calculator and see which one matches the equation.

A practical suggestion!
This is the problem with multiple choice.

As for the question:
[MATH]\cos(\frac{\pi}{2}-x)=\sin(x)\\ \sin(\frac{\pi}{2}-x)=\cos(x)\\ \cos(-x)=\cos(x)[/MATH]
[MATH]\frac{\frac{\cos^2(\frac{\pi}{2}-x)}{\sin^2(\frac{\pi}{2}-x)}-\frac{1}{\cos^2(-x)}}{\frac{1}{\cos x}-\frac{\sin^2x}{\cos x}}\\ \text{ }\\ =\dfrac{\frac{\sin^2x}{\cos^2x}-\frac{1}{\cos^2x}}{\frac{1}{\cos x}-\frac{\sin^2x}{\cos x}}[/MATH]
Multiply top and bottom by [MATH]\hspace2ex \cos^2x \hspace2ex[/MATH] and tidy up.

 

[nasi112]

A practical suggestion!
This is the problem with multiple choice.

As for the question:
[MATH]\cos(\frac{\pi}{2}-x)=\sin(x)\\ \sin(\frac{\pi}{2}-x)=\cos(x)\\ \cos(-x)=\cos(x)[/MATH]
[MATH]\frac{\frac{\cos^2(\frac{\pi}{2}-x)}{\sin^2(\frac{\pi}{2}-x)}-\frac{1}{\cos^2(-x)}}{\frac{1}{\cos x}-\frac{\sin^2x}{\cos x}}\\ \text{ }\\ =\dfrac{\frac{\sin^2x}{\cos^2x}-\frac{1}{\cos^2x}}{\frac{1}{\cos x}-\frac{\sin^2x}{\cos x}}[/MATH]
Multiply top and bottom by [MATH]\hspace2ex \cos^2x \hspace2ex[/MATH] and tidy up.

Beautiful Solution.

 

[Deleted member 4993]

A practical suggestion! This is the problem with multiple choice.....

Original statement is complicated compound functions of "cot" and "sec". Most of the calculators - that I know of - do not have "one button" evaluation of those functions. The time required to input the correct expression would be considerable - fraught with multiple chance of mistakes. You test for at least two values of "x" to be reasonably sure about the answer.

 

[lex]

Original statement is complicated compound functions of "cot" and "sec". Most of the calculators - that I know of - do not have "one button" evaluation of those functions. The time required to input the correct expression would be considerable - fraught with multiple chance of mistakes. You test for at least two values of "x" to be reasonably sure about the answer.

I wouldn't see the lack of a button for cot and sec as being a major problem. It only requires one button press to fix: [MATH]\boxed{\hspace1ex x^{-1} \hspace1ex}[/MATH]However I do agree, it is a complicated expression and plenty of potential for errors!
Secondly in situations when I test expressions like this for correctness, I simply use an 'obscure value', like 0.27 and if they match up, you can be pretty sure they match up!
I think it is a worthwhile tactic to know. I have used it many times before to check the correctness of algebraic manipulations I have done.
In fact that is a more realistic use in this situation - checking your answer after you have done the algebra.
(However if for some reason I couldn't get the algebra done, I have to confess I would be only too happy to try this method to attempt to get the mark)!