Simplifying this trig solution

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The expression csc(x)sqrt(sec^2(x)-1) can be simplified using the Pythagorean identity 1 + tan^2(x) = sec^2(x). By substituting this identity, the expression simplifies to csc(x)tan(x). This further simplifies to 1/cos(x), which equals sec(x). The restriction of 0 ≤ x < π/2 ensures that all trigonometric functions are non-negative, allowing the square root to be simplified without ambiguity. The final simplified expression is sec(x).
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Homework Statement


Simplify this expression: csc(x)sqrt(sec^2(x)-1) with 0 less than or equal x less than pi/2.


Homework Equations





The Attempt at a Solution


Not really sure what they want here, but would a good place to start be to square the entire expression to get rid of the square root or would that make things more complicated? and what is meant by x being between 0 and pi/2?
 
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jhahler said:

Homework Statement


Simplify this expression: csc(x)sqrt(sec^2(x)-1) with 0 less than or equal x less than pi/2.

Homework Equations



The Attempt at a Solution


Not really sure what they want here, but would a good place to start be to square the entire expression to get rid of the square root or would that make things more complicated? and what is meant by x being between 0 and pi/2?
Squaring the expression is not recommended.

Do you know a trig identity relating tan2(x) and sec2(x) ?

Use that.
 
gotcha, the Pythagorean identity 1+tan^2(x)=sec^2(x)
 
ok, I think I got this if anyone wants to confirm, still not sure about the x greater or equal to 0 and less than or equal to pi/2, but here goes. csc(x)sqrt(sec^2(x)-1) turns into csc(x)sqrt(tan^2(x)+1-1) after the Pythagorean identity, then simplifies to csc(x)tan(x) after you do the square root. csc(x)tan(x) is equal to 1/sin(x) * 1/cos(x), those multiply together to give 1/cos(x) and that equals sec(x), that sound right?
 
jhahler said:
ok, I think I got this if anyone wants to confirm, still not sure about the x greater or equal to 0 and less than or equal to pi/2, but here goes. csc(x)sqrt(sec^2(x)-1) turns into csc(x)sqrt(tan^2(x)+1-1) after the Pythagorean identity, then simplifies to csc(x)tan(x) after you do the square root. csc(x)tan(x) is equal to 1/sin(x) * 1/cos(x), those multiply together to give 1/cos(x) and that equals sec(x), that sound right?

I think you mean \frac{1}{sinx}\times\frac{sinx}{cosx}.

But yes, that looks right.
 
jhahler said:
ok, I think I got this if anyone wants to confirm, still not sure about the x greater or equal to 0 and less than or equal to pi/2, but here goes. csc(x)sqrt(sec^2(x)-1) turns into csc(x)sqrt(tan^2(x)+1-1) after the Pythagorean identity, then simplifies to csc(x)tan(x) after you do the square root. csc(x)tan(x) is equal to 1/sin(x) * 1/cos(x), those multiply together to give 1/cos(x) and that equals sec(x), that sound right?
The restriction: 0 ≤ x < π/2 means that all of the trig functions give a non-negative result and in particular, \sqrt{\tan^2(x)=\tan(x)} without the ± sign.
 
yes, bread18, i did mean \frac{sinx}{\cos(x)}

Thanks by the way!
 
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