If asinx=bcosx=ctanx=k, thenbc+ck1+1+bkak is

sinx=akcosx=bktanx=ckSince,tanx=cosxsinx,ck= ba Or, bc= kack=tanx⇒ck1=cotx⇒ 1+bkak = 1+cosxsinx⇒ ck1+ 1+bkak=cotx+ 1+cosxsinx= sinx(1+cosx)cosx+cos2x+sin2x=sinx(1+cosx)1+cosx= cosecxHence, bc+ck1+1+bkak = ka+ak1

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If { x}{a}={ x}{b}={ x}{c}=k, then bc+ {1}{ck}+{ak}{1+bk} is

Question

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

k (a + \frac{1}{a})

\frac{a}{k} + \frac{1}{ak}

\frac{1}{k ^{2}}

^{\frac{a}{k}}

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Olivia Discussed

\frac{sin x}{a} = \frac{cos x}{b} = \frac{tan x}{c} = k

, then

b c + \frac{1}{c k} + \frac{ak}{1 + bk}

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

s in x = ak

cos x = bk

t an x = c k

Since,

t an x = \frac{s in x}{cos x}

c k = \frac{a}{b}

Or,

b c = \frac{a}{k}

c k = t an x

\Rightarrow \frac{1}{c k} = co t x

\Rightarrow \frac{ak}{1 + bk} = \frac{s in x}{1 + cos x}

\Rightarrow \frac{1}{c k} + \frac{ak}{1 + bk} = co t x + \frac{s in x}{1 + cos x}

= \frac{cos x + cos ^{2} x + s in ^{2} x}{s in x ( 1 + cos x )}

= \frac{1 + cos x}{s in x ( 1 + cos x )}

= cosec x

Hence,

b c + \frac{1}{c k} + \frac{ak}{1 + bk}

\frac{a}{k} + \frac{1}{ak}

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Question Text	If $\frac{sin x}{a} = \frac{cos x}{b} = \frac{tan x}{c} = k$ , then $b c + \frac{1}{c k} + \frac{ak}{1 + bk}$ is
Answer Type	Text solution:1
Upvotes	150

If asinx​=bcosx​=ctanx​=k, thenbc+ck1​+1+bkak​ is

Text solutionVerified

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