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STEP I 2007 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2007 question 2 solution

A = \tan ^{-1} \left( \frac{1}{2} \right) \implies \tan{A} = \frac{1}{2}

B = \tan ^{-1} \left( \frac{1}{3} \right) \implies \tan{B} = \frac{1}{3}

Hence:

\tan{ \left( A + B \right) } = \frac{ \tan{A} + \tan{B} }{1 - \tan{A} \tan{B}} = \frac{ \frac{1}{2} + \frac{1}{3} }{ 1 - \frac{1}{6}} = 1

Hence:

A + B = \tan ^{-1} (1) = \frac{ \pi }{4} .


Consider:

A = \tan ^{-1} \left( \frac{1}{p} \right) \implies \tan{A} = \frac{1}{p}

B = \tan ^{-1} \left( \frac{1}{q} \right) \implies \tan{B} = \frac{1}{q}

A + B = \frac{ \pi }{4} \implies \tan{ \left( A + B \right) } = 1

Hence:

1 = \frac{ \frac{1}{p} + \frac{1}{q} }{1 - \frac{1}{pq}} = \frac{ \frac{ p + q }{pq} }{ \frac{ pq - 1 }{pq} } = \frac{p + q}{pq - 1}

Hence:

pq - 1 = p + q \implies pq - p - q = 1 \implies (p - 1)(q - 1) - 1 = 1 \implies (p - 1)(q - 1) = 2 .

Hence, as "p", and "q" are positive integers, p = 3 , q = 2 (or p = 2 , q = 3 ).

A = \tan ^{-1} \left( \frac{1}{r} \right) \implies \tan{A} = \frac{1}{r}

B = \tan ^{-1} \left( \frac{s}{s + t} \right) \implies \tan{B} = \frac{s}{s + t} .

In the aforementioned context:

p = r

q = \frac{s + t}{s}

Hence, for the stated (on the examination paper) equation to be true:

(r - 1) \left( \frac{s + t}{s} - 1 \right) = 2

Hence:

(r - 1) \left( \frac{t}{s} \right) = 2 .

As (t, \ s) = 1 (given), the fraction of \frac{t}{s} is in lowest terms.

t(r - 1) = 2s .

Consider that r \in \mathbb{Z} \implies (r - 1) \in \mathbb{Z} :

\left( \frac{2s}{t} \right) \in \mathbb{Z}

Hence, as s \in \mathbb{Z} , \left( \frac{2}{t} \right) \in \mathbb{Z} , and as t \in \mathbb{Z} there can be two possible values of "t":

t = 1, \ or \ 2

t = 1 \implies r - 1 = 2s \implies r = 2s + 1

t = 2 \implies 2r - 2 = 2s \implies r = s + 1 .

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