It so happens that all 4 problems are easy and "short" to solve/verify, otherwise you should have posted them separtately. Asking for many problems in only one posting is too much if the problems are "hard" or "long" to solve.Originally Posted bylizzy20

If one doesn't have time to solve all of the problems, he might ignore all of the problems outright, and you'd get no answer from him. Or if it is me, even if I have enough time, if I cannot solve all of the problems, I will ignore all of them and I will bypass your question.

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[cscX] /[cot +tanX] = cosX

This kind, we convert all functions into their sine or cosine equivalents.

Lefthand Side, (LFS) =

= [cscX] /[cotX +tanX]

= [1 /sinX] /[(cosX/sinX) +(sinX/cosX)]

= [1 /sinX] /[(cosX*cosX +sinX*sinX) /(sinX*cosX)]

= [1 /sinX] /[(cos2(X) +sin^2(x)) /(sinXcosX)]

Since sin^2(X) +cos^2(x) = 1, then,

= [1 /sinX] /[1 /(sinXcosX)]

= [1 /sinX]*[(sinXcosX) /1]

= cosX

= Righthand Side, (RHS)

Verified.

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[1 /(sinY -1)] -[1 /(sinY +1)] = -2sec^2(Y)

LFS =

= [1 /(sinY -1)] -[1 /(sinY +1)]

Combine them into one fcraction only,

= [(sinY +1)*1 -(sinY -1)*1] /[(sinY -1)(sinY +1)]

= [sinY +1 -sinY +1] /[(sinY)^2 -1^2]

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

From sin^2(Y) +cos^2(Y) = 1, we can get sin^2(Y) = 1 -cos^2(Y), so,

= 2 /[(1 -cos^2(Y)) -1]

= 2 /[-cos^2(y)]

= -2sec^2(Y)

= RHS

Verified.

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sin^3(X) -cos^3(X) = (1 +sinXcosX)(sinX -cosX)

The LHS is a difference of cubes, in the form a^3 -b^3. Factoring that,

a^3 -b^3

= (a-b)(a^2 +ab +b^2)

So, adopting that to the LHS,

= (sinX -cosX)[sin^2(X) +sinXcosX +cos^2(X)]

= (sinX -cosX)[sin^2(X) +cos^2(X) +sinXcosX]

= (sinX -cosX)[1 +sinXcosX]

= (1 +sinXcosX)(sinX -cosX)

= RHS

Verified.

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[tan(theta)] +[(cos(theta)) /(1 +sin(theta))] = sec(theta)

Let X = theta, for less confusion, and easy typing.

[tanX] +[cosX /(1 +sinX)] = secX

LHS =

= tanX +[cosX /(1 +sinX)]

= [sinX /cosX] +[cosX /(1 +sinX)]

= [(1 +sinX)*sinX +cosX*cosX] /[cosX*(1 +sinX)]

= [sinX +sin^2(X) +cos^2(X)] /[cosX*(1 +sinX)]

= [sinX +1] /[cosX*(1 +sinX)]

The (sinX +1) or (1 +sinX) is called out,

= [1] /[cosX]

= secX

= RHS

Verified.

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I assume you know how to combine two fractions into one fraction only.

a/b +c/d

The common denominator is b*d,

= (d*a +b*c) /(b*d).