4 Prove that \(\sec ^ { 2 } \theta + \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta \operatorname { cosec } ^ { 2 } \theta\).
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Question 4:
Answer Marks
Guidance
Answer Mark
Guidance
LHS \(= \sec^2\theta + \cosec^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta}\) M1
Use of \(\sec\theta = 1/\cos\theta\) and \(\cosec\theta = 1/\sin\theta\), not just stating\(= \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta\sin^2\theta}\) M1
Adding
\(= \frac{1}{\cos^2\theta\sin^2\theta}\) A1
Use of \(\cos^2\theta + \sin^2\theta = 1\) soi
\(= \sec^2\theta\cosec^2\theta\) A1
AG OR \(\sec^2\theta + \cosec^2\theta = \tan^2\theta + 1 + \cot^2\theta + 1 = \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2\)M1
Correct formulae oe
\(= \frac{\cos^4\theta + \sin^4\theta + 2\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta}\) M1
Adding
\(= \frac{(\cos^2\theta + \sin^2\theta)^2}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta} = \sec^2\theta\cosec^2\theta\) A1, A1
Use of Pythagoras; AG
OR working with both sides: LHS \(\sec^2\theta + \cosec^2\theta = \tan^2\theta + 1 + \cot^2\theta + 1 = \tan^2\theta + \cot^2\theta + 2\)M1
Correct formulae used on one side
RHS \(= (1+\tan^2\theta)(1+\cot^2\theta) = 1 + \tan^2\theta + \cot^2\theta + \tan^2\theta\cot^2\theta\) M1
Use of same formulae on other side
\(= \tan^2\theta + \cot^2\theta + 2 =\) LHS A1
Use of \(\tan\theta\cot\theta = 1\) oe, dependent on both method marks
A1 Showing equal
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## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| LHS $= \sec^2\theta + \cosec^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta}$ | M1 | **Use** of $\sec\theta = 1/\cos\theta$ and $\cosec\theta = 1/\sin\theta$, not just stating |
| $= \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta\sin^2\theta}$ | M1 | Adding |
| $= \frac{1}{\cos^2\theta\sin^2\theta}$ | A1 | Use of $\cos^2\theta + \sin^2\theta = 1$ soi |
| $= \sec^2\theta\cosec^2\theta$ | A1 | **AG** |
| **OR** $\sec^2\theta + \cosec^2\theta = \tan^2\theta + 1 + \cot^2\theta + 1 = \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2$ | M1 | Correct formulae oe |
| $= \frac{\cos^4\theta + \sin^4\theta + 2\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta}$ | M1 | Adding |
| $= \frac{(\cos^2\theta + \sin^2\theta)^2}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta} = \sec^2\theta\cosec^2\theta$ | A1, A1 | Use of Pythagoras; **AG** |
| **OR** working with both sides: LHS $\sec^2\theta + \cosec^2\theta = \tan^2\theta + 1 + \cot^2\theta + 1 = \tan^2\theta + \cot^2\theta + 2$ | M1 | Correct formulae used on one side |
| RHS $= (1+\tan^2\theta)(1+\cot^2\theta) = 1 + \tan^2\theta + \cot^2\theta + \tan^2\theta\cot^2\theta$ | M1 | Use of same formulae on other side |
| $= \tan^2\theta + \cot^2\theta + 2 =$ LHS | A1 | Use of $\tan\theta\cot\theta = 1$ oe, dependent on both method marks |
| | A1 | Showing equal |
| | **[4]** | |
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4 Prove that $\sec ^ { 2 } \theta + \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta \operatorname { cosec } ^ { 2 } \theta$.
\hfill \mbox{\textit{OCR MEI C4 Q4 [4]}}