prove the following trig identities

Question

sec(θ) - sin(θ)tan(θ) = cos(θ)

1 - sin(t)/1 + sin(t) = (sec(t) - tan(t))²

sin(x) + cos(x) = 1 + cot(x)/csc(x)

Answer 1

If sec(θ) - sin(θ)tan(θ) = cos(θ), then substituting sec(θ) = 1/cos(θ) and tan(θ) = sin(θ)/cos(θ) gives

1/cos(θ) - sin²(θ)/cos(θ) = cos(θ). Now multiply both sides by cos(θ) to get

1 - sin²(θ) = cos²(θ). Now adding sin²(θ) to both sides leads to the known identity 1 = sin²(θ) + cos²(θ)

If (1 - sin(t))/(1 + sin(t)) = (sec(t) - tan(t))² = (1/cos(t) - sin(t)/cos(t))² = (1 - sin(t))²/cos²(t)

Divide both sides by 1 - sin(t) to get

1/(1 + sin(t)) = (1 - sin(t))/cos²(t). Now multiply both sides by 1 + sin(t) get

1 = (1 + sin(t))(1 - sin(t))/cos²(t) = (1 - sin²(t))/cos²(t). Now multiply both sides by cos²(θ) to get

cos²(θ) = 1 - sin²(θ). Now adding sin²(θ) to both sides leads to the known identity cos²(θ) + sin²(θ) = 1

sin(x) + cos(x) = (1 + cot(x))/csc(x) = (1 + cos(x)/sin(x))/(1/sin(x))

Multiplying both numerator and denominator of the right hand side (RHS) by sin(x) gives

RHS = (1 + cos(x)/sin(x))/(1/sin(x)) = (sin(x) + cos(x))/1 = sin(x) + cos(x) which is the same as the left hand side, proving the identity.

Answer 2

Recall that cos²x + sin²x = 1. So, cos²x = 1 - sin²x and sin²x = 1 - cos²x.

(sect - tant)² = (1/ cost - sint / cost)² = (1 - sint)² / cos²t = (1 - sint)² / (1 - sin²t)

Factor the denominator and simplify.

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secθ - sinθtanθ = (1/cosθ) - sin²θ / cosθ

Combine the fractions and simplify.

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(1 + cotx) / cscx = (1 + cosx/sinx) ÷ (1/sinx) = (sinx / sinx + cosx/sinx) ÷ (1/sinx)

Simplify.