Simplifying this trig solution

[jhahler]

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?

 

[SammyS]

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.

 

[jhahler]

gotcha, the Pythagorean identity 1+tan^2(x)=sec^2(x)

 

[jhahler]

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?

 

[Bread18]

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.

 

[SammyS]

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.

 

[jhahler]

yes, bread18, i did mean $$\frac{sinx}{\cos(x)}$$

Thanks by the way!