AS June 2022 Q3

EdexcelAS paperCurrent spec7 markst-Formulae

(a) Use \(t = \tan\dfrac{\theta}{2}\) to show that, where both sides are defined\[\frac{29 - 21\sec\theta}{20 - 21\tan\theta} \equiv \frac{5t + 2}{2t + 5}\] (4)

(b) Hence, again using \(t = \tan\dfrac{\theta}{2}\), prove that, where both sides are defined\[\frac{20 + 21\tan\theta}{29 + 21\sec\theta} \equiv \frac{29 - 21\sec\theta}{20 - 21\tan\theta}\] (3)

Mark scheme (a)

Scheme	Marks	AO
Sight of \(\sec\theta = \dfrac{1 + t^2}{1 - t^2}\) and \(\tan\theta = \dfrac{2t}{1 - t^2}\) at least once each.	B1	1.2
\(\dfrac{29 - 21\sec\theta}{20 - 21\tan\theta} = \dfrac{29 - 21\left(\dfrac{1 + t^2}{1 - t^2}\right)}{20 - 21\left(\dfrac{2t}{1 - t^2}\right)}\)	M1	1.1b
\(= \dfrac{29(1 - t^2) - 21(1 + t^2)}{20(1 - t^2) - 21(2t)} = \dfrac{8 - 50t^2}{-20t^2 - 42t + 20}\) or \(= \dfrac{\left(\dfrac{29(1 - t^2) - 21(1 + t^2)}{(1 - t^2)}\right)}{\left(\dfrac{20(1 - t^2) - 21(2t)}{(1 - t^2)}\right)} = \dfrac{8 - 50t^2}{-20t^2 - 42t + 20}\)	M1	2.1
\(= \dfrac{-2(5t - 2)(5t + 2)}{-2(5t - 2)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\ \) cso \(= \dfrac{2(2 + 5t)(2 - 5t)}{2(2 - 5t)(5 + 2t)} = \dfrac{5t + 2}{2t + 5}\ \) cso	A1*	1.1b
	(4)

Notes

B1: Uses the correct identities at least once each. May be seen in (a) or (b)

M1: Substitutes their identities into the equation, they need not be correct

M1: Multiplies numerator and denominator by \(1 - t^2\) and simplifies to a quadratic with all terms collected
Special case: If they make an error with the identities award M1 if they use the correct method to write the numerator and denominator as a single fraction and then divides to achieve an expression of the form \(\dfrac{a}{b}\)

A1*: Cancels factors to achieve the given expression, with no errors seen
Condone \(\dfrac{50t^2 - 8}{20t^2 + 42t - 20} = \dfrac{(5t + 2)(5t - 2)}{(5t - 2)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\) and \(\dfrac{-8 + 50t^2}{-20t^2 - 42t + 20} = \dfrac{(5t + 2)(5t - 2)}{(5t - 2)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\)

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{20 + 21\tan\theta}{29 + 21\sec\theta} = \dfrac{20 + 21\left(\dfrac{2t}{1 - t^2}\right)}{29 + 21\left(\dfrac{1 + t^2}{1 - t^2}\right)}\) \(= \dfrac{20(1 - t^2) + 21(2t)}{29(1 - t^2) + 21(1 + t^2)}\)	M1	2.1
\(\dfrac{20 + 42t - 20t^2}{50 - 8t^2} = \dfrac{2(5 - 2t)(5t + 2)}{2(5 - 2t)(5 + 2t)} = \ldots\) \(\dfrac{20 + 42t - 20t^2}{50 - 8t^2} = \dfrac{-2(2t - 5)(5t + 2)}{-2(2t + 5)(2t - 5)} = \ldots\)	M1	1.1b
Achieves form correct working \(= \dfrac{5t + 2}{2t + 5}\ *\) Then concludes hence the result is true or = RHS or \(= \dfrac{5t + 2}{2t + 5} = \dfrac{29 - 21\sec\theta}{20 - 21\tan\theta}\) or \(\dfrac{20 + 21\tan\theta}{29 + 21\sec\theta} = \dfrac{29 - 21\sec\theta}{20 - 21\tan\theta}\) LHS = RHS	A1*	2.2a
	(3)
	(7 marks)

Notes

M1: Substitutes into LHS of equation and multiplies numerator and denominator by \(1 - t^2\) to work towards the other side.

M1: Simplifies to a quadratic with all terms collected and factorises then cancels terms in the LHS in an attempt to try and match expressions.

A1*: Achieves correct expressions for both sides and gives a conclusion deducing that the result is true.

M1A0 for \(\dfrac{20t^2 - 42t - 20}{8t^2 - 50} = \dfrac{(5t + 2)(2t - 5)}{(2t - 5)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\) and
\(\dfrac{-20t^2 - 42t + 20}{-8 + 50t^2} = \dfrac{(5t + 2)(2t - 5)}{(2t - 5)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\)