5 Prove that \(\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }\).
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Question 5:
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Guidance
\(\text{LHS} = \cot\beta - \cot\alpha = \dfrac{\cos\beta}{\sin\beta} - \dfrac{\cos\alpha}{\sin\alpha}\) M1
\(\cot = \cos/\sin\)
\(= \dfrac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\sin\alpha\sin\beta}\) M1
Combining fractions
\(= \dfrac{\sin(\alpha-\beta)}{\sin\alpha\sin\beta}\) E1
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OR (from RHS): \(\dfrac{\sin(\alpha-\beta)}{\sin\alpha\sin\beta} = \dfrac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\sin\alpha\sin\beta}\)M1
Using compound angle formula
\(= \dfrac{\cos\beta}{\sin\beta} - \dfrac{\cos\alpha}{\sin\alpha} = \cot\beta - \cot\alpha\) M1, E1
Splitting fractions; using \(\cot = \cos/\sin\)
[3]
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## Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{LHS} = \cot\beta - \cot\alpha = \dfrac{\cos\beta}{\sin\beta} - \dfrac{\cos\alpha}{\sin\alpha}$ | M1 | $\cot = \cos/\sin$ |
| $= \dfrac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\sin\alpha\sin\beta}$ | M1 | Combining fractions |
| $= \dfrac{\sin(\alpha-\beta)}{\sin\alpha\sin\beta}$ | E1 | www |
| **OR (from RHS):** $\dfrac{\sin(\alpha-\beta)}{\sin\alpha\sin\beta} = \dfrac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\sin\alpha\sin\beta}$ | M1 | Using compound angle formula |
| $= \dfrac{\cos\beta}{\sin\beta} - \dfrac{\cos\alpha}{\sin\alpha} = \cot\beta - \cot\alpha$ | M1, E1 | Splitting fractions; using $\cot = \cos/\sin$ |
| **[3]** | | |
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5 Prove that $\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }$.
\hfill \mbox{\textit{OCR MEI C4 Q5 [3]}}