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Trigonometry, 2ADV T2 2025 HSC 22

Prove that

\(\dfrac{\sin ^4 \theta+\cos ^4 \theta}{\sin ^2 \theta\, \cos ^2 \theta}+2=\sec ^2 \theta\, \operatorname{cosec}^2 \theta\).   (2 marks)

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\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{Note: RHS } =\sec ^2 \theta\, \operatorname{cosec}^2 \theta=\dfrac{1}{\sin ^2 \theta\, \cos ^2 \theta}\)

\(\text{RHS}\) \(=\dfrac{\sin ^4 \theta+\cos ^4 \theta}{\sin ^2 \theta \cos ^2 \theta}+2\)
  \(=\dfrac{\sin ^4 \theta+\cos ^4 \theta+2 \sin ^2 \theta \cos ^2 \theta}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\dfrac{\left(\sin ^2 \theta+\cos ^2 \theta\right)^2}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\dfrac{1}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\sec ^2 \theta\, \operatorname{cosec} ^2 \theta\)

Trigonometry, 2ADV T2 EQ-Bank 1

Given  \(\tan \theta=\dfrac{3}{2}\)  and  \(0°<\theta<90°\),

find the value of  \(\dfrac{1-\sin (180-\theta)}{\cos (180+\theta)}\).   (3 marks)

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\(\dfrac{1-\sqrt{5}}{2}\)

Show Worked Solution

\(\text{Since} \ \ \tan \theta=\dfrac{1}{2}: \)
 

\(\dfrac{1-\sin (180-\theta)}{\cos (180+\theta)}\) \(=\dfrac{1-\sin \theta}{-\cos \theta}\)
  \(=\dfrac{\sin \theta-1}{\cos \theta}\)
  \(=\dfrac{\frac{1}{\sqrt{5}}-1}{\frac{2}{\sqrt{5}}} \times \dfrac{\sqrt{5}}{\sqrt{5}}\)
  \(=\dfrac{1-\sqrt{5}}{2}\)

Trigonometry, 2ADV T2 EQ-Bank 2 MC

Given  \(\tan \theta=\cfrac{1}{3}\)  and  \(0°<\theta<90°\),

find the value of  \(\dfrac{1-\sin (180+\theta)}{\cos (90-\theta)}\).

  1. \(\sqrt{10}+1\)
  2. \(1\)
  3. \(\sqrt{10}-1\)
  4. \(\dfrac{\sqrt{10}}{2}\)
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\(\Rightarrow A\)

Show Worked Solution

\(\text {Since} \ \ \tan \theta=\dfrac{1}{3}:\)
 

\(\dfrac{1-\sin (180+\theta)}{\cos (90-\theta)}\) \(=\dfrac{1+\sin \theta}{\sin \theta}\)
  \(=\dfrac{1+\frac{1}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} \times \dfrac{\sqrt{10}}{\sqrt{10}}\)
  \(=\sqrt{10}+1\)

 

\(\Rightarrow A\)

Trigonometry, 2ADV T2 SM-Bank 4

Solve the equation  \(2 \cos 2 \theta=2 \sin 2 \theta\)  for  \(0 \leqslant \theta \leqslant 2 \pi\)   (2 marks)

Show Answers Only

\(\theta=\displaystyle\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}, \frac{13 \pi}{8} \)

Show Worked Solution
\(2 \cos 2 \theta\) \(=2 \sin 2 \theta\)
\(\dfrac{2 \sin 2 \theta}{2 \cos 2 \theta}\) \(=1\)
\(\tan 2 \theta\) \(=1\)

\(2 \theta\) \(=\displaystyle \frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \frac{13 \pi}{4}, \cdots\)
\(\theta\) \(=\displaystyle\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}, \frac{13 \pi}{8} \quad(0 \leqslant \theta \leqslant 2 \pi)\)

Trigonometry, 2ADV T2 EQ-Bank 7

Prove  \(\dfrac{1+\cot \theta}{1+\tan \theta}=\cot \theta\).   (3 marks)

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  \(\text{LHS}\) \(=\dfrac{1+\dfrac{\cos \theta}{\sin \theta}}{1+\dfrac{\sin \theta}{\cos \theta}}\)
    \(=\dfrac{\dfrac{\sin \theta+\cos \theta}{\sin \theta}}{\dfrac{\cos \theta+\sin \theta}{\cos \theta}}\)
    \(=\dfrac{\sin \theta+\cos \theta}{\sin \theta} \times \dfrac{\cos \theta}{\cos \theta+\sin \theta}\)
    \(=\dfrac{\cos \theta}{\sin \theta}\)
    \(=\cot \theta\)
    \(\ =\text{ RHS}\)
Show Worked Solution
  \(\text{LHS}\) \(=\dfrac{1+\dfrac{\cos \theta}{\sin \theta}}{1+\dfrac{\sin \theta}{\cos \theta}}\)
    \(=\dfrac{\dfrac{\sin \theta+\cos \theta}{\sin \theta}}{\dfrac{\cos \theta+\sin \theta}{\cos \theta}}\)
    \(=\dfrac{\sin \theta+\cos \theta}{\sin \theta} \times \dfrac{\cos \theta}{\cos \theta+\sin \theta}\)
    \(=\dfrac{\cos \theta}{\sin \theta}\)
    \(=\cot \theta\)
    \(\ =\text{ RHS}\)

Trigonometry, 2ADV T2 EQ-Bank 6

Simplify  \(\sin \,  \theta \, \cos \theta \,\operatorname{cosec}^2 \,  \theta\).   (2 marks)

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\(\cot \theta\)

Show Worked Solution

\(\sin \theta \cos \theta \times \dfrac{1}{\sin ^2 \theta}\)

\(=\dfrac{\cos \, \theta}{\sin \, \theta}\)

\(=\cot \, \theta\)

Trigonometry, 2ADV T2 EQ-Bank 5

Solve  \(\sin x-\cos x=0 \quad-\pi \leqslant x \leqslant \pi\)   (2 marks)

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\(x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)

Show Worked Solution
  \(\sin x-\cos x\) \(=0\)
  \(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\) \(=0\)
  \(\tan x-1\) \(=0\)
  \(\tan x\) \(=1\)
  \(x\) \(=\tan ^{-1}(1)\)

 
\(\therefore x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)

Trigonometry, 2ADV T2 EQ-Bank 4

Prove  \(\dfrac{\operatorname{cosec} \theta+\sec \theta}{1+\tan \theta}=\operatorname{cosec} \theta\).   (3 marks)

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\(\text{Prove:}\ \ \dfrac{\operatorname{cosec} \theta+\sec \theta}{1+\tan \theta}=\operatorname{cosec} \theta\)

  \(\text{LHS}\) \(= \dfrac{\operatorname{cosec} \theta + \sec \theta}{1 + \tan \theta}\)
    \(=\dfrac{\dfrac{1}{\sin \theta} + \dfrac{1}{\cos \theta}}{1 + \dfrac{\sin \theta}{\cos \theta}} \times \dfrac{\cos \theta}{\cos \theta} \)
    \(=\dfrac{\dfrac{\cos \theta}{\sin \theta}+1}{\cos \theta+\sin \theta}\)
    \(=\dfrac{\dfrac{\cos \theta+\sin \theta}{\sin \theta}}{\cos \theta+\sin \theta}\)
    \(=\dfrac{1}{\sin \theta}\)
    \(=\operatorname{cosec} \theta \quad \text{… as required.}\)
Show Worked Solution

\(\text{Prove:}\ \ \dfrac{\operatorname{cosec} \theta+\sec \theta}{1+\tan \theta}=\operatorname{cosec} \theta\)

  \(\text{LHS}\) \(= \dfrac{\operatorname{cosec} \theta + \sec \theta}{1 + \tan \theta}\)
    \(=\dfrac{\dfrac{1}{\sin \theta} + \dfrac{1}{\cos \theta}}{1 + \dfrac{\sin \theta}{\cos \theta}} \times \dfrac{\cos \theta}{\cos \theta} \)
    \(=\dfrac{\dfrac{\cos \theta}{\sin \theta}+1}{\cos \theta+\sin \theta}\)
    \(=\dfrac{\dfrac{\cos \theta+\sin \theta}{\sin \theta}}{\cos \theta+\sin \theta}\)
    \(=\dfrac{1}{\sin \theta}\)
    \(=\operatorname{cosec} \theta \quad \text{… as required.}\)

Trigonometry, 2ADV T2 EQ-Bank 2

Express  \(3 \operatorname{cosec}(180+x)+5 \cos (90-x)\)  as a single fraction in terms of \(\sin x\), given all angles are measured in degrees.   (3 marks)

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\(\dfrac{-3+5 \sin ^2 x}{\sin x}\)

Show Worked Solution

\(3 \operatorname{cosec}(180+x)+5 \cos (90-x)\)

\(=\dfrac{3}{\sin \left(180^{\circ}+x\right)}+5 \sin x\)

\(=\dfrac{3}{-\sin x}+5 \sin x\)

\(=\dfrac{-3}{\sin x}+\dfrac{5 \sin ^2 x}{\sin x}\)

\(=\dfrac{-3+5 \sin ^2 x}{\sin x}\)

Trigonometry, 2ADV T2 2020 HSC 19

Prove that  `sec theta-cos theta = sin theta\ tan theta.`   (2 marks)

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`text(LHS)` `=1/cos theta-cos theta`  
  `=(1-cos^2 theta)/cos theta`  
  `=sin^2 theta/cos theta`  
  `=sin theta * sin theta/cos theta`  
  `=sin theta\ tan theta\ …\ text(as required)`  
Show Worked Solution
`text(LHS)` `=1/cos theta-cos theta`  
  `=(1-cos^2 theta)/cos theta`  
  `=sin^2 theta/cos theta`  
  `=sin theta * sin theta/cos theta`  
  `=sin theta\ tan theta\ …\ text(as required)`  

Trigonometry, 2ADV T2 SM-Bank 2

Find all solutions of the equation  `2 cos theta = sqrt 3 cot theta`,  for  `0<=theta<=2pi`   (3 marks)

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`theta=pi/3, pi/2, (2pi)/3, (3pi)/2`

Show Worked Solution

`2 cos theta = sqrt 3 cot theta`

`2 cos theta-sqrt 3 cot theta` `= 0`  
`2 cos theta-sqrt 3 (cos theta)/(sin theta)` `=0`  
`(2-sqrt 3/sin theta) cos theta` `=0`  

 
`text(If)\ \ cos theta=0,`

`theta=pi/2, (3pi)/2`
  

`text(If)\ \ 2-sqrt 3/sin theta = 0\ \ =>\ \ sin theta = sqrt 3/2`

`theta = pi/3, (2pi)/3`

Trigonometry, 2ADV T2 SM-Bank 44 MC

The domain of the function with rule  `f(x) = 1 - sec(x + pi/4)`  is

  1. `text(all real)\ x`
  2. `{((4k - 1)pi)/4}, (text{for}\ k\ text{integer}) `
  3. `{((4k + 1)pi)/4}, (text{for}\ k\ text{integer}) `
  4. `{((2k - 1)pi)/4}, (text{for}\ k\ text{integer}) `
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`C`

Show Worked Solution

`y = sec (x)=1/cos(x)\ \ text(has asymptotes when)`

`x = −pi/2, pi/2, (3pi)/2, …`

`=> y = sec (x + pi/4)\ \ text(has asymptotes at when)`

`x = (−3pi)/4, pi/4, (5pi)/4, …`
 

`:.\ text(Domain:)\ {((4k + 1)pi)/4}, (text{for}\ k\ text{integer}) `

`=>C`

Trigonometry, 2ADV T2 2019 HSC 13a

Solve  `2 sin x cos x = sin x`  for  `0 <= x <= 2pi`.  (3 marks)

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`x = 0, quad pi/3, quad pi, quad (5 pi)/3`

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♦ Mean mark 49%.

`2 sin x cos x-sin x` `= 0`
`sin x (2 cos x-1)` `= 0`
`sin x` `= 0`
`=> x` `= 0,\  pi,\  2pi`
`cos x` `= 1/2`
`=> x` `= pi/3, (5 pi)/3`

 
`:. x = 0, quad pi/3, quad pi, quad (5 pi)/3,quad 2pi`

Trigonometry, 2ADV T2 SM-Bank 42

Prove that

`(1 - sin^2 x cos^2 x)/(sin^2 x) = cot^2 x + sin^2 x`.  (2 marks)

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`text(See Worked Solutions)`

Show Worked Solution
`text(RHS)` `= (cos^2 x)/(sin^2 x) + sin^2 x`
  `= (cos^2 x + sin^4 x)/(sin^2 x)`
  `= (cos^2 x + sin^2 x(1 – cos^2 x))/(sin^2 x)`
  `= (cos^2 x + sin^2 x – sin^2 x cos^2 x)/(sin^2 x)`
  `= (1 – sin^2 x cos^2 x)/(sin^2 x)`
  `= \ text(LHS)`

Trigonometry, 2ADV T2 SM-Bank 41

Prove that

`(secx + tanx)(secx - tanx) = 1`.  (2 marks)

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`text(See Worked Solutions)`

Show Worked Solution
`text(LHS)` `= (secx + tanx)(secx – tanx)`
  `= sec^2x – tan^2x`
  `= 1/(cos^2x) – (sin^2 x)/(cos^2 x)`
  `= (1 – sin^2 x)/(cos^2 x)`
  `= (cos^2 x)/(cos^2 x)`
  `= 1`
  `=\ text(RHS)`

Trigonometry, 2ADV T2 SM-Bank 40

Let  `(tantheta-1) (sin theta-sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`.

  1. State all possible values of  `tan theta`.  (1 mark)

  2. Hence, find all possible solutions for  `(tan theta-1) (sin^2 theta-3 cos^2 theta) = 0`, where  `0 <= theta <= pi`.  (2 marks)

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  1. `tan theta = 1 or tan theta = +- sqrt 3`
  2. `theta = pi/4, pi/3 or (2 pi)/3`
Show Worked Solution

i.  `(tantheta-1) (sin theta-sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`

`=> tan theta = 1`

♦ Mean mark 42%.
`=>sin theta-sqrt 3 cos theta` `=0`
`sin theta` `=sqrt3 cos theta`
`tan theta` `=sqrt3`

 

`=>sin theta + sqrt 3 cos theta` `=0`
`sin theta` `=-sqrt3 cos theta`
`tan theta` `=-sqrt3`

 
`:. tan theta = 1 or tan theta = +- sqrt 3`

 

ii.  `(tan theta-1) (sin^2 theta-3 cos^2 theta) = 0`

`text(Using part a:)`

♦ Mean mark 42%.

`(tan theta-1) (sin theta-sqrt 3 cos theta) (sin theta + sqrt 3 cos theta) = 0`

`=> tan theta` `= 1` `qquad or qquad` `tan theta` `= +- sqrt 3`
`theta` `= pi/4`   `theta` `= pi/3, (2 pi)/3`

 
`:. theta = pi/4, pi/3 or (2 pi)/3\ \ \ \ (0<=theta<=pi)`

Trigonometry, 2ADV T2 SM-Bank 34

Solve the equation  `sqrt 3 sin x = cos x`  for  `– pi<=x<= pi`.   (2 marks)

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`x = pi/6,\ \ \ – (5 pi)/6`

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`text(Divide both sides by)\ cos x :`

MARKER’S COMMENT: Many students who found the base angle correctly could not solve within the restrictions.
`sqrt 3 sin x` `=cos x`
`sqrt 3 tan x` `= 1`
`tan x` `= 1/sqrt 3`
`=>\ text(Base angle)\ = pi/6`

 
`:. x = pi/6\ \ text(or)\ -(5 pi)/6,\ \ \ (-pi <=x<= pi)`

Trigonometry, 2ADV T2 SM-Bank 32

Express  `5cot^2 x - 2text(cosec)\ x + 2`  in terms of  `text(cosec)\ x`  and hence solve

`5cot^2 x - 2text(cosec)\ x + 2 = 0`  for  `0 < x < 2pi`.  (3 marks)

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`x = pi/2`

Show Worked Solution
`cot^2 x` `= (cos^2 x)/(sin^2 x)`
  `= (1 – sin^2 x)/(sin^2 x)`
  `= text(cosec)^2 x – 1`

 

`5cot^2 x – 2text(cosec)\ x + 2` `= 0`
`5(text(cosec)^2 x – 1) – 2text(cosec)\ x + 2` `= 0`
`5text(cosec)^2 x – 2text(cosec)\ x – 3` `= 0`
`(5text(cosec)\ x + 3)(text(cosec)\ x – 1)` `= 0`
`text(cosec)\ x` `= −3/5` `text(cosec)\ x` `= 1`
`sinx` `= −5/3` `sinx` `= 1`
`(text(no solution))` `x` `= pi/2`

 
`:. x = pi/2`

Trigonometry, 2ADV T2 2017 HSC 7 MC

Which expression is equivalent to  `tan theta + cot theta`?

  1. `text(cosec)\ theta + sec theta`
  2. `sec theta\ text(cosec)\ theta`
  3. `2`
  4. `1`
Show Answers Only

`B`

Show Worked Solution
`tan theta + cot theta` `= (sin theta)/(cos theta) + (cos theta)/(sin theta)`
  `= (sin^2 theta + cos^2 theta)/(cos theta sin theta)`
  `= 1/(cos theta sin theta)`
  `= sec theta\ text(cosec)\ theta`

`=>  B`

Trigonometry, 2ADV T2 2016 HSC 8 MC

How many solutions does the equation  `|\ cos (2x)\ | = 1`  have for  `0 <= x <= 2 pi?`

  1. `1`
  2. `3`
  3. `4`
  4. `5` 
Show Answers Only

`D`

Show Worked Solution

`|\ cos (2x)\ | = 1`

♦♦♦ Mean mark 23%.

`cos (2x) = +- 1`

`text(When)\ \ cos (2x)` `= 1`
`2x` `= 0, 2pi, 4 pi, …`
`:. x` `= 0, pi, 2 pi, …`

 

`text(When)\ \ cos (2x)` `= – 1`
`2x` `= pi, 3 pi, 5 pi, …`
`:. x` `= pi/2, (3 pi)/2, (5 pi)/2, …`

 

`:. x = 0, pi/2, pi, (3 pi)/2, 2 pi\ \ \ text(for)\ \ \ 0 <= x <= 2pi`

`=>  D`

Trigonometry, 2ADV T2 2004 HSC 9a

Consider the geometric series  `1 − tan^2 theta + tan^4 theta − …`

  1. When the limiting sum exists, find its value in simplest form.  (2 marks)

  2. For what values of  `theta`  in the interval
     
        `−pi/2 < theta < pi/2`  does the limiting sum of the series exist?  (2 marks)

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  1. `cos^2 theta`
  2. `− pi/4 < theta < pi/4`
Show Worked Solution

i.   `1 − tan^2 theta + tan^4 theta − …`

`=>\ text(GP where)\ \ a=1,\ \ r=T_2/T_1= − tan^2 theta`

`:. S_∞` `= 1/(1 − (−tan^2 theta))`
  `= 1/(1 + tan^2 theta)`
  `= 1/(sec^2 theta)`
  `= cos^2 theta`

 

ii.   `text(Find)\ \ theta\ \ text(such that)\ \ \ |\ r\ |` `< 1`
  `|−tan^2 theta\ |` `< 1`
  ` tan^2 theta` `< 1`
  `−1 < tan theta` `< 1`
  `:. − pi/4 < theta` `< pi/4`

Trigonometry, 2ADV T2 2004 HSC 8a

  1. Show that  `cos theta tan theta = sin theta`.   (1 mark)

  2. Hence solve  `8 sin theta cos theta tan theta = text(cosec)\ theta`  for  `0 ≤ theta ≤ 2pi`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `pi/6, (5pi)/6`
Show Worked Solution

i.    `text(Prove)\ \ cos theta tan theta = sin theta`

`text(LHS)` `= cos theta tan theta`
  `= cos theta ((sin theta)/(cos theta))`
  `= sin theta`
  `=\ text{RHS}`

 

ii.    `8 sin theta cos theta tan theta` `= text(cosec)\ theta`
   `:. 8 sin theta(sin theta)` `= text(cosec)\ theta,\ \ \ \ text{(part (i))}` 
  `8 sin^2 theta`  `= 1/(sin theta)` 
  `8 sin^3 theta`  `= 1` 
  `sin^3 theta`  `= 1/8` 
  `sin theta`  `= 1/2` 
   `:. theta` `= pi/6, (5pi)/6\ \ \ \ text{(for}\ \ 0 ≤ theta ≤ 2pi text{)}` 

Trigonometry, 2ADV T2 2014 HSC 15a

Find all solutions of  `2 sin^2 x + cos x − 2 = 0`, where  `0 <= x <= 2pi`.   (3 marks)

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`x = pi/3,\ pi/2,\ (3pi)/2,\ (5pi)/3`

Show Worked Solution
♦ Mean mark 42%
`2 sin^2 x + cos x\ – 2` `= 0`
`2(1\ – cos^2x) + cos x\ – 2` `= 0`
`2\ – 2cos^2x + cosx\ – 2` `= 0`
`-2cos^2x + cosx` `= 0`
`cosx (-2 cosx + 1)` `= 0`

 

`:. -2 cosx + 1` `= 0` `\ text(or)\ \ \ \ \ \ \ ` `cos x` `= 0`
`2 cos x` `= 1`   `x` `= pi/2,\ (3pi)/2`
`cos x` `= 1/2`      
`cos(pi/3)` `=1/2`      

 

`text(S)text(ince cos is positive in)\ 1^text(st) // 4^text(th)\ text(quadrants,)`

`x` `= pi/3,\ 2 pi \  – pi/3`
  `= pi/3,\ (5pi)/3`

 

`:. x = pi/3,\ pi/2,\ (3pi)/2,\ (5pi)/3\ \ text(for)\ \ 0 <= x <= 2pi`

Trigonometry, 2ADV T2 2014 HSC 7 MC

How many solutions of the equation  `(sin x-1)(tan x + 2) = 0`  lie between  `0`  and  `2 pi`?

  1. `1`
  2. `2`
  3. `3`
  4. `4`
Show Answers Only

`B`

Show Worked Solution
♦♦♦ Mean mark 25%, making it the toughest MC question in the 2014 exam.
COMMENT: Note that the “2 solutions” answer relies on the sum of an infinity of zeros not equalling zero. This concept created unintended difficulty in this question.

`text(When)\ (sin x-1)(tan x + 2) = 0`

`(sinx-1) = 0\ \ text(or)\ \ tan x + 2 = 0`

`text(If)\ \ sin x-1= 0:`

`sin x= 1\ \ =>\ \ x= pi/2,\ \ \ 0 < x < 2 pi`
 

`text(If)\ \ tan x + 2= 0:`

`tan x= -2`

`text{Since}\ tan\ pi/2\ text{is undefined, there are only 2 solutions when}`

`tan x = -2\ \text{(which occurs in the 1st and 4th quadrants).}`
  

`:.\ 2\ text(solutions)`

`=>  B`

Calculus, 2ADV C4 2010 HSC 5b

  1. Prove that  `sec^2 x + secx tanx = (1 + sinx)/(cos^2x)`.   (1 mark)

  2. Hence prove that  `sec^2 x + secx tanx = 1/(1 - sinx)`.     (1 mark)

  3. Hence, use the identity  `int sec ax tan ax\ dx=1/a sec ax`  to find the exact value of

     

          `int_0^(pi/4) 1/(1 - sinx)\ dx`.   (2 marks)

Show Answers Only
  1. `text(Proof)  text{(See Worked Solutions)}`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `sqrt2`
Show Worked Solution

i.    `text(Need to prove)`

`sec^2x + secxtanx = (1 + sinx)/(cos^2x)`
 

`text(LHS)` `=sec^2x + secx tanx`
  `=1/(cos^2x) + 1/(cosx) xx (sinx)/cosx`
  `=1/(cos^2x) + (sinx)/(cos^2x)`
  `=(1 + sinx)/(cos^2x)`
  `= text(RHS)\ \ \ \ text(… as required)`

 

ii.   `text(Need to prove)` 

♦♦ Mean mark 31%.
`sec^2x + secx tanx` `= 1/(1\ – sinx)`
`text(i.e.)\ \ (1 + sinx)/(cos^2x)` `= 1/(1\ – sin x)\ \ \ \ \ text{(part (i))}`
`text(LHS)` `= (1 + sinx)/(cos^2x)`
  `=(1 + sin x)/(1\ – sin^2x)`
  `=(1 + sinx)/((1\ – sinx)(1 + sinx)`
  `=1/(1\ – sinx)\ \ \ \ text(… as required)`

 

iii.  `int_0^(pi/4) 1/(1\ – sinx)\ dx`

♦ Mean mark 37%.

`= int_0^(pi/4) (sec^2x + secx tanx)\ dx`

`= [tanx + secx]_0^(pi/4)`

`= [(tan(pi/4) + sec(pi/4)) – (tan0 + sec0)]`

`= [(1 + 1/(cos(pi/4)))\ – (0 + 1/(cos0))]`

`= 1 + sqrt2\ – 1`

`= sqrt2`