If sin x-a-cos x-b-tan x-c-k- then bc -1-c k-a2-1-b k is equal toA- 1-ka-1-aB- ka-1-aC- 1-k2D- a-k

The correct option is A cos xtan x/k^2 + 1/tan x + sin x/1 + cos x = sin x/k^2 + cos x( 1 + cos x) + sin^2x/sin x( 1 + cos x) = a/k + 1/sin x = a/k + 1/ak ...

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Standard XI

Mathematics

Sin2a and Cos2a in Terms of Tana

If sin x/a=co...

Question

If $\frac{sin x}{a} = \frac{cos x}{b} = \frac{tan x}{c} = k$ , then $b c + \frac{1}{c k} + \frac{a k}{1 + b k}$ is equal to

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Solution

The correct option is B
$\frac{cos x tan x}{k^{2}} + \frac{1}{tan x} + \frac{sin x}{1 + cos x} = \frac{sin x}{k^{2}} + \frac{cos x (1 + cos x) + {sin}^{2} x}{sin x (1 + cos x)} = \frac{a}{k} + \frac{1}{sin x} = \frac{a}{k} + \frac{1}{a k}$

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Similar questions

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If sinxa=cosxb=tanxc=k, then bc+1ck+ak1+bk is equal to

The correct option is B cosxtanxk2+1tanx+sinx1+cosx=sinxk2+cosx(1+cosx)+sin2xsinx(1+cosx)=ak+1sinx=ak+1ak

If $\frac{sin x}{a} = \frac{cos x}{b} = \frac{tan x}{c} = k$ , then $b c + \frac{1}{c k} + \frac{a k}{1 + b k}$ is equal to

The correct option is B
$\frac{cos x tan x}{k^{2}} + \frac{1}{tan x} + \frac{sin x}{1 + cos x} = \frac{sin x}{k^{2}} + \frac{cos x (1 + cos x) + {sin}^{2} x}{sin x (1 + cos x)} = \frac{a}{k} + \frac{1}{sin x} = \frac{a}{k} + \frac{1}{a k}$