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If f(theta)=|sin^2A cot A1sin^2BcosB1sin...

If `f(theta)=|sin^2A cot A1sin^2BcosB1sin^2CcosC1|` , then `t a n A+t a n B+c` `cotAcotBcotC` `sin^2A+sin^2B+sin^2C` 0

A

` tan A + tan B+C`

B

`cot A cot B cot C`

C

`sin^(2) A + sin^(2) B+ sin^(2) C `

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
D
Applying `R_(2) to R_(2)- R_(1) " and " R_(3) to R_(3) -R_(1)` we get
`Delta =|{:(sin^(2)A,,cot A,,1),(sin (B+A) sin(B-A),,(sin(A-B))/(sinA sinB),,0),(sin(C +A) sin (C-A),,(sin(A-C))/(sin A sin C),,0):}|`
`[ :' cot alpha -cot beta = (sin(Beta -alpha))/(sin alpha sin beta)]`
Expanding along `C_(3)` we get
`Delta =(sin (A-B) sin (A-C))/(sin A) [-(sin (B+A))/(sin C) +(sin (C+A))/(sinB)]`
`=(sin (A-B)sin(A-C))/(sin A) [-(sin (pi-C))/(sinC)+(sin(pi-beta))/(sin B)]`
`=(sin (A-B) sin (A -C))/(sinA) [-(sin C)/(sin C)+(sinB)/(sinB)]=0`
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