1. prove that cosθ/(1-tanθ)=(cosθ-1)/(1-〖sec〗^2 θ)?

Question

1. prove that cosθ/(1-tanθ)=(cosθ-1)/(1-〖sec〗^2 θ)

2. simplify tanθ- cosθ/(1-sinθ)

3. find the equation of the tangent to the curve y=〖4x〗^2+
2x+sin x at x= n/3

Answer 1

Something odd seems to have happened with the first question, but it looks like it should be:
prove that cos θ / (1 - tan θ) = (cos θ - 1) / (1 - sec^2 θ)
Unfortunately, this isn't true; consider θ = π/6, then cos θ = √3/2, sin θ = 1/2, tan θ = 1/√3, sec θ = 2/√3.
The LHS is (√3/2) / (1 - 1/√3) = (3/2) / (√3 - 1) = 3 / [2(√3 - 1)] ≈ 2.05.
The RHS is (√3 / 2 - 1) / (1 - 4/3) = (3 - 3√3/2) ≈ 0.402.

I can prove that cos θ / (1 - tan θ) = (cos θ - 1) / [(1 - tan θ) (1 - sec θ)] = (cos θ - 1) / (1 - tan θ - sec θ + tan θ sec θ). {To see this, cross-multiply and simplify - you should get cos θ - sin θ - 1 + tan θ on both sides.} This doesn't seem very close to what you've written, though.

2. tan θ - (cos θ / (1 - sin θ))
= tan θ - [cos θ (1 + sin θ)] / [(1 - sin θ) (1 + sin θ)]
= tan θ - [cos θ (1 + sin θ)] / (1 - sin^2 θ)
= tan θ - [cos θ (1 + sin θ)] / cos^2 θ
= tan θ - sec θ - tan θ
= - sec θ.

3. I assume you mean x = π/3 in the last line.
y = (4x)^2 + 2x + sin x
y'(x) = 2(4x).4 + 2 + cos x
= 32x + 2 + cos x.
y'(π/3) = 32 (π/3) + 2 + cos (π/3)
= 32π/3 + 5/2.
y(π/3) = (4π/3)^2 + 2π/3 + sin (π/3)
= 16/9 π^2 + 2/3 π + √3 / 2.
So the equation is, rather messily,
y - (16/9 π^2 + 2/3 π + √3 / 2) = (32π/3 + 5/2) (x - π/3)
= (32π/3 + 5/2) x - 32π^2/9 - 5π/6
so y = (32π/3 + 5/2) x - 32π^2/9 - 5π/6 + (16/9 π^2 + 2/3 π + √3/2)
= (32π/3 + 5/2) x - 16π^2/9 - π/6 + √3/2.

Answer 2

i think of i ought to respond to extensive type 3 and four! Tan ^4 ? + tan²? observe: sec²? = tan²? + a million sec ^4 ? = tan ^4 ? + 2tan²? + a million = (sec ^4 ? - 2tan²? - a million) + (sec²? - a million) = (sec ^4 ? - 2(sec²? - a million) - a million) + (sec²? - a million) = sec ^4 ? - (sec²? - a million) - a million = sec ^4 ? - sec²? Cos^4 ? - sin^4 ? = cos²? – sin²? = (cos²? – sin²?)(cos²? + sin²?) observe: cos²? + sin²? = a million (pythagorean id) = cos²? – sin²? there you bypass. i'm going to be coming back to characteristic solutions if I easily have the luxurious of time. i desire this enables! ;)