Anonymous
Anonymous asked in Science & MathematicsMathematics · 11 months ago

# cscx - sinx = cotx*cosx?

with steps please? ive been trying to figure it out to no luck

Relevance

To prove that csc(x) - sin(x) = cot(x) * cos(x), we begin evaluating from the LHS to show that it's equal to the RHS, i.e.

LHS:

csc(x) - sin(x)

= [1/sin(x)] - sin(x) {from the property: csc(x) = 1/sin (x)}

= [1 - sin²(x)]/sin(x)

= cos²(x)/sin (x) {from the Pythagorean Property: sin²(x) + cos²(x) = 1}

= [cos(x)/sin(x)] * cos(x) {from the concept: cos²(x) = cos(x) * cos(x)}

= cot(x) * cos(x)

= RHS

• Prove Cscx - sinx = cotx*cosx

There really is no specific step for trig identities. You just have to toy with it using your knowledge. The point of practicing these is to get your familiar with trig identities by getting you to apply them.

(1/sin(x))-(sin(x))=(cos(x)/sin(x))(cos(x))

Finding common denominator for the left side

(1/sin(x))-(sin^2(x)/sin(x))=(cos^2(x)/sin(x))

The rest can be done simply by understanding cos^2(x)+sin^2(x)=1

If you get cos^2(x) by itself you get cos^2(x)=1-sin^2(x)

You have that as a numerator at the left side so just substitute and you get

cos^2(x)/sin(x)=cos^2(x)/sin(x) and you are done there

• H
Lv 7
11 months agoReport

And those three dots seem to be blocking the full thing.

• It’s all trigonometry to me.

• csc(x) - sin(x) (left)

= 1/sin(x) - sin(x)

= 1/sin(x) - sin^2(x)/sin(x)

= (1 - sin^2(x)) / sin(x)

= cos^2(x)/sin(x)

= (cos(x)/sin(x))cos(x)

= cot(x)cos(x) (right). Done.

You got it right. :)

• LHS:

csc (x) - sin (x)

= 1/sin(x) - sin(x)

= 1/2 (csc(x) + cos(2 x) csc(x))

= 1/2 (cos(2 x) + 1) csc(x)

= cos(x) cot(x)

= RHS

cscx= 1/sinx

1/sinx -sinx= (1-sin^2x)/ sinx = cos^2x/sinx

we know that cotx= cosx/sinx

cotx * cosx = (cosx/sinx) cosx = cos^2x/sinx

Hence, proved

• 1/sin x - sin x

1 - sin² x

---------------------

sin x

cos²x

------------ = cot x cos x

sin x

• RHS:

cot(x) * cos(x)

= [cos(x) / sin(x)] * cos(x)

= cos^2(x) / sin(x)

= [1 - sin^2(x)] / sin(x) # Pythagorean Trigonometric identity:

= [1 / sin(x)] - [sin^2(x) / sin(x)]

= csc(x) - sin(x)

• LHS

= csc(x) - sin(x)

= 1/sin(x) - sin²(x)/sin(x)

= [1 - sin²(x)]/sin(x)

= cos²(x)/sin(x)

= [cos(x)/sin(x)]cos(x)

= cot(x)cos(x)

= RHS