Suppose θ and (ϕ=0) are such that sec(θ+ϕ), secθ and sec(θ−ϕ) are in A.P. If cosθ=kcos(2ϕ) for some k, then k is equal to?

sec(θ+ϕ)+sec(θ−ϕ)=2secθcos(θ+ϕ)cos(θ−ϕ)cos(θ+ϕ)+cos(θ−ϕ)=cosθ2cos2θcos2ϕ2cosθcosϕ=cosθ1cos2θcosϕ=cos2θ+cos2ϕ−1cos2θ(cosϕ−1)=cos2ϕ−1cos2θ=cosϕ+1cos2θ=2cos22ϕcosθ=2cos2ϕk=2

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Suppose and ( 0) are such that ( +), and ( -) are in A.P. If

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Suppose $θ$ and $(ϕ \neq = 0)$ are such that $sec (θ + ϕ)$ , $sec θ$ and $sec (θ - ϕ)$ are in A.P. If $cos θ = k cos (\frac{ϕ}{2})$ for some k, then k is equal to?

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Suppose

θ

and

(ϕ \neq = 0)

are such that

sec (θ + ϕ)

sec θ

and

sec (θ - ϕ)

are in A.P. If

cos θ = k cos (\frac{ϕ}{2})

for some k, then k is equal to?

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13 mins ago

Text solutionVerified

sec (θ + ϕ) + sec (θ - ϕ) = 2 sec θ

\frac{cos ( θ + ϕ ) + cos ( θ - ϕ )}{cos ( θ + ϕ ) cos ( θ - ϕ )} = \frac{2}{cos θ}

\frac{2 cos θ cos ϕ}{cos 2 θ cos 2 ϕ} = \frac{1}{cos θ}

cos^{2} θ cos ϕ = cos^{2} θ + cos^{2} ϕ - 1

cos^{2} θ (cos ϕ - 1) = cos^{2} ϕ - 1

cos^{2} θ = cos ϕ + 1

cos^{2} θ = 2 cos^{2} \frac{ϕ}{2}

cos θ = 2 cos \frac{ϕ}{2}

k = 2

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Question Text	Suppose $θ$ and $(ϕ \neq = 0)$ are such that $sec (θ + ϕ)$ , $sec θ$ and $sec (θ - ϕ)$ are in A.P. If $cos θ = k cos (\frac{ϕ}{2})$ for some k, then k is equal to?
Answer Type	Text solution:1
Upvotes	150

Suppose θ and (ϕ=0) are such that sec(θ+ϕ), secθ and sec(θ−ϕ) are in A.P. If cosθ=kcos(2ϕ​) for some k, then k is equal to?

Text solutionVerified

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Suppose $θ$ and $(ϕ \neq = 0)$ are such that $sec (θ + ϕ)$ , $sec θ$ and $sec (θ - ϕ)$ are in A.P. If $cos θ = k cos (\frac{ϕ}{2})$ for some k, then k is equal to?