Prove Identities: (cscAcotA-secAtanA)/(cosA-sinA) = (1+cscASecA)/(sinAcosA)?

Answer 1

[csc(A)cot(A) - sec(A)tan(A)] / [cos(A) - sin(A)] =
[1 + csc(A)sec(A)] / [sin(A)cos(A)]

To solve this we use the more complex side. That would be the left hand side.

LHS = [csc(A)cot(A) - sec(A)tan(A)] / [cos(A) - sin(A)]

Change everything to sines and cosines.

LHS = [ (1/sinA) (cosA/sinA) - (1/cosA) (sinA/cosA)] / [cosA - sinA]

Multiply out each fraction.

LHS = [ ( cosA/sin^2(A) ) - ( sinA / cos^2(A) ) ] / [cosA - sinA]

To get rid of this complex fraction, multiply everything by
sin^2(A)cos^2(A)

LHS = [ cosA cos^2(A) - sinA sin^2(A) ] / [cosA - sinA] [sin^2(A)cos^2(A)]

Simplify the top. We should have cosine cubed and sine cubed.

LHS = [cos^3(A) - sin^3(A)] / [cosA - sinA][sin^2(A) cos^2(A)]

Note that the numerator is a difference of cubes. We know how to factor difference of cubes. a^3 - b^3 = (a - b)(a^2 + ab + b^2)

LHS = [cosA - sinA] [cos^2(A) + sinAcosA + sin^2(A)] / ([cosA - sinA][sin^2(A) cos^2(A)])

Now how the cosA - sinA on the top and bottom are common; cancel them out. This leaves us with

LHS = [cos^2(A) + sinAcosA + sin^2(A)] / [sin^2(A) cos^2(A)]

Look how, on the numerator, we have cos^2(A) + sin^2(A). This is equal to 1.

LHS = [1 + sinAcosA] / [sin^2(A) cos^2(A)]

Let's split this up into two fractions.

LHS = 1/[sin^2(A) cos^2(A)] + sinAcosA / [sin^2(A) cos^2(A)]

We can change the first fraction into a product of two fractions. Also, stuff cancels in the second fraction.

LHS = [1/sin^2(A)] [1/cos^2(A)] + 1/[sinAcosA]

By definition,

LHS = [csc^2(A) sec^2(A)] + 1/[sinAcosA]

Now, bring them under a common denominator.

LHS = { [csc^2(A) sec^2(A)] [sinAcosA] + 1 } / (sinAcosA)

LHS = { [1/sin^2(A)][1/cos^2(A)] sinAcosA + 1 } / (sinAcosA)

Note the cancellations on the numerator.

LHS = [ (1/sinA) (1/cosA) + 1 ] / (sinAcosA)

By definition,

LHS = [ cscA secA + 1 ] / (sinAcosA)

Addition is commutative; swap the terms in the numerator.

LHS = [ 1 + cscA secA ] / (sinAcosA) = RHS

Answer 2

Comes close to to the "Bent Air regulation" of Wind power Aerodynamics, which relates the stress stepped forward on a piece of wind turbine blade from the conventional attitude of deflection of the great mass of wind flowing interior of reach brought about by utilising it. (understand that air is heavy, vast in great parts - many an awful lot in line with 2d close to a protracted blade.) that's F = dCV^2 ( a million - cos a ) the place d is the air density and C is a relentless of proportionality. This "regulation" may well be derived from Newton's regulation utilising the trig id 2 sin^2 (a) = (a million - cos 2a). T-Shirts would quickly be obtainable with the Bent Air regulation on them and that i'm hoping that is no longer direct mail in pointing out this.

Answer 3

Prove the identity:
(cscAcotA-secAtanA)/(cosA-sinA) = (1+cscASecA)/(sinAcosA)

We will start with the left hand side.

(cscAcotA - secAtanA)/(cosA - sinA)
= {(1/sinA)(cosA/sinA) - (1/cosA)(sinA/cosA)}/(cosA - sinA)
= (cos³A - sin³A)/{(sin²A)(cos²A)(cosA - sinA)}
= {(cosA - sinA)(cos²A + (cosA)(sinA) + sin²A)}
/{(sin²A)(cos²A)(cosA - sinA)}
= {(cosA)(sinA) + (cos²A + sin²A)}/{(sin²A)(cos²A)}
= {(cosA)(sinA) + 1}/{(sin²A)(cos²A)}
= {1 + 1/[(cosA)(sinA)]}/{(sin²A)(cos²A)}
= {1 + (secA)(cscA)}/{(sinA)(cosA)}
= {1 + (cscA)(secA)}/{(sinA)(cosA)} = Right Hand Side