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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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\(\text{See Worked Solutions}\)

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

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

Trigonometry, 2ADV T2 EQ-Bank 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 4

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

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

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

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

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)`  
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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)`  

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)`

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`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)`

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`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 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`
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`B`

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`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 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)

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

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)

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  1. `text(Proof)  text{(See Worked Solutions)}`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `sqrt2`
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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`