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If `(sinx)/a=(cosx)/b=(tanx)/c=k ,` then `b c+1/(c k)+(a k)/(1+b k)` is equal to `k(a+1/a)` (b) `1//k(a+1/a)` `1/(k^2)` (d) `a/k`
A. `k(a+1/a)`
B. `1/k(a+1/a)`
C. `1/k^2`
D. `a/k`

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Best answer
Correct Answer - B
`bc+1/(ck)+(ak)/(1+bk)=(cosxtanx)/k^2+1/tanx+sinx/(1+cosx)`
`=sinx/k^2+(cosx(1+cosx)+sin^2x)/(sinx(1+cosx))`
`a/k+(cosx(1+cosx)+(1-cos^2x))/(sinx(1+cosx))`
`=a/k+1/sinx`
`a/k+1/(ak)`

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