# {(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

*print*Print*list*Cite

### 1 Answer

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

``