Definitions, graphs, and identities involving sec, cosec, and cot.
Answer at least 3 of 3 correctly to complete this section.
At AS level you worked with sin, cos, and tan. At A2, you meet their reciprocals: sec, cosec, and cot. These are not new functions — they are shorthand for cos1, sin1, and tan1 (or sincos). But they come with their own identities, graphs, and derivatives that appear frequently in A2 exam questions.
The two new Pythagorean identities (1+tan2θ=sec2θ and 1+cot2θ=cosec2θ) are particularly important — they are used in integration, differential equations, and proof questions.
secθ=cosθ1cosecθ=sinθ1cotθ=sinθcosθ=tanθ1
Each is undefined where its "parent" function equals zero:
| Function | Undefined when | Values | |----------|---------------|--------| | secθ | cosθ=0 | θ=2π,23π,… | | cosecθ | sinθ=0 | θ=0,π,2π,… | | cotθ | sinθ=0 (tanθ undefined) | θ=0,π,2π,… |
Critical distinction: secθ is the reciprocal of cosθ, meaning secθ=cosθ1. It is not the inverse function cos−1θ (which gives you the angle whose cosine is θ). The notation is confusing, but the meaning is completely different.
| Angle | sec | cosec | cot | |-------|--------|----------|--------| | 6π | 32 | 2 | 3 | | 4π | 2 | 2 | 1 | | 3π | 2 | 32 | 31 |
These follow directly from the exact values of sin, cos, and tan — just take the reciprocal.
Find the exact value of sec3π+cosec6π.
sec3π=cos3π1=211=2
cosec6π=sin6π1=211=2
sec3π+cosec6π=2+2=4
Answer at least 3 of 3 correctly to complete this section.
Start from sin2θ+cos2θ=1 and divide through:
Divide by cos2θ:
cos2θsin2θ+1=cos2θ1⟹tan2θ+1=sec2θ
Divide by sin2θ:
1+sin2θcos2θ=sin2θ1⟹1+cot2θ=cosec2θ
Memory aid: tan pairs with sec (they share the same asymptotes on the unit circle). cot pairs with cosec.
y=secθ is the reciprocal of y=cosθ:
y=cosecθ is the reciprocal of y=sinθ:
y=cotθ is the reciprocal of y=tanθ:
Simplify sec2θ−tan2θ.
Using 1+tan2θ=sec2θ:
sec2θ−tan2θ=1
This is always true (provided secθ is defined).
Solve 2sec2θ−5tanθ−5=0 for 0≤θ≤2π.
Replace sec2θ=1+tan2θ:
2(1+tan2θ)−5tanθ−5=0
2+2tan2θ−5tanθ−5=0
2tan2θ−5tanθ−3=0
Let t=tanθ: (2t+1)(t−3)=0
t=−21ort=3
When tanθ=−21: α=tan−1(21)≈0.4636
Tangent is negative in the 2nd and 4th quadrants:
θ=π−0.4636≈2.678 or θ=2π−0.4636≈5.820
When tanθ=3: α=tan−1(3)≈1.2490
Tangent is positive in the 1st and 3rd quadrants:
θ≈1.249 or θ=π+1.249≈4.391
Solutions: θ≈1.25,2.68,4.39,5.82 (to 2 d.p.)
Prove that cosec2θ−cosecθcotθ=1+cosθ1.
Start with the LHS. Write everything in terms of sinθ and cosθ:
LHS=sin2θ1−sinθ1×sinθcosθ=sin2θ1−sin2θcosθ
=sin2θ1−cosθ
Replace sin2θ=1−cos2θ=(1−cosθ)(1+cosθ):
=(1−cosθ)(1+cosθ)1−cosθ=1+cosθ1=RHS□
Answer at least 3 of 4 correctly to complete this section.
These are derived using the chain rule (or quotient rule) applied to the reciprocal definitions.
dxd(secx)=secxtanx
dxd(cosecx)=−cosecxcotx
dxd(cotx)=−cosec2x
Write secx=(cosx)−1 and apply the chain rule:
dxd(cosx)−1=−1×(cosx)−2×(−sinx)
=cos2xsinx=cosx1×cosxsinx=secxtanx
Write cosecx=(sinx)−1:
dxd(sinx)−1=−1×(sinx)−2×cosx
=−sin2xcosx=−sinx1×sinxcosx=−cosecxcotx
Watch the minus sign: secx differentiates to a positive expression, but cosecx differentiates to a negative one. Similarly, cotx differentiates to −cosec2x (negative), while tanx differentiates to +sec2x (positive). The "co-" functions pick up a minus sign.
Differentiate y=3sec2x.
Using the chain rule:
dxdy=3×sec2xtan2x×2=6sec2xtan2x
Find dxdy when y=x2cotx.
Using the product rule with u=x2 and v=cotx:
dxdy=2xcotx+x2×(−cosec2x)=2xcotx−x2cosec2x
The curve C has equation y=2secx−4cosx for 0<x<2π. Find the exact coordinates of the stationary point.
dxdy=2secxtanx+4sinx
Set equal to zero:
2secxtanx+4sinx=0
Write in terms of sin and cos:
cos2x2sinx+4sinx=0
2sinx(cos2x1+2)=0
Since 0<x<2π, sinx=0, and cos2x1+2>0 always. So this equation has no solution.
Let us instead consider a classic exam example:
Given y=secx+cosecx, find dxdy and show that when dxdy=0 in (0,2π), tan3x=1.
dxdy=secxtanx−cosecxcotx
Set equal to zero:
secxtanx=cosecxcotx
cos2xsinx=sin2xcosx
Cross-multiply: sin3x=cos3x
cos3xsin3x=1⟹tan3x=1⟹tanx=1⟹x=4π
Answer at least 3 of 3 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.