If `A , B , C , D` are the angles of a quadrilateral, then `(tanA+tanB+tanC+tanD)/(cotA+cotB+cotC+cotD)i se q u a lto:` `tanAtanBtanCtanD` `cotAcotBcotCcotD` `tan^2A+tan^2B+tan^2C+tan^2D` `sumtanAtanBtanC`

Options

A. cot A cot B cot C cot D
B. tan A tan B tan C tan D
C. `-tan A tan B tan C tan D`
D. `-cot A cot B cot C cot D`

Answer

B

Explanation

`A+B+C+D=2pi`
`therefore A+B=2pi-(C+D)`
`therefore tan(A+B)=tan [2pi - C+D)]=-tan(C+D)`
`rArr (tan A + tan B)/(1-tan A tan B)=-(tan C + tan D)/(1-tan C tan D)`
`rArr (tan A+ tan B)(1-tan C tan D)`
`=-(1-tan A tan B)(tan C+tan D)`
`rArr tan A+tan B-tan A tan C tan D?-tan B tan C tan D`
`=-[(tan C+tan D-tan A tan B tan C-tan A tan B tan D)]`
`rArr tan A + tan B+ tan C+tan D`
`= tan A tan B tan C + tan A tan C tan D+ tan A tan D + tan B tan C tan D`
Dividing both sides by tan A tan B tan C tan D, we get
`(tan A+ tan B + tan C + tan D)/(ta A tan B tan C tan D)`
`=(1)/(tan D)+(1)/(tan D)+(1)/(tan C)+(1)/(tan A)`
`= cot D + cot B + cot C + cot A`
`rArr (tan A +tan B + tan C + tan D)/(cot A + cot B + cot C + cot D) = tan A tan B tan C tan D`