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{(tanu - tanv)/(1 + tanu.tanv)}{(1 + cotu.cotv)/(cotu - cotv)} equals to: a.(sinu -...

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nasirjam | Student, Grade 9 | (Level 2) Honors

Posted April 7, 2013 at 7:08 AM via web

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{(tanu - tanv)/(1 + tanu.tanv)}{(1 + cotu.cotv)/(cotu - cotv)} equals to:

a.(sinu - sinv)/(1 + sinu.sinv)

b.-1

c.(sinv - sinu)/(1 + sinu.sinv)

d.1 + tanu.tanv

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted April 7, 2013 at 7:57 AM (Answer #1)

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Ans. -1

Since

{(tan(u)-tan(v))/(1+tan(u)tan(v))}= tan(u-v) (i)

{(1+cot(u).cot(v))/(cot(u)-cot(v))}={(tan(u).tan(v)+1)/(tan(u).tan

(v))}/{(tan(v)-tan(u))/(tan(u)tan(v))}

=1/tan(v-u) (ii)

multiply (i) and (ii) we have

tan(u-v)/tan(v-u)=-1

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