Suppose θ and φ(≠ 0) are such that (θ+φ) , sec θ and (θ-φ) are in A.P. If θ=k ((φ)/(2)) for some k , then k is equal to

Correct answer: A. ± 2 Since, (θ-φ), θ and (θ+φ) are in A.P., aligned & 2 θ= (θ-φ)+ (θ+φ) & (2)/( θ)=( (θ+φ)+ (θ-φ))/( (θ-φ) (θ+φ)) aligned aligned & 2( ^2 θ- ^2 φ)= θ[2 θ φ & ^2 θ(1- φ)= ^2 φ=1- ^2 φ & ^2 θ=1+ φ=2 ^2 (φ)/(2) & θ=± 2 (φ)/(2) & But given θ=k (φ)/(2) & k=± 2 aligned

JEE Main 201219 MayOnline

Exam: JEE Main
Year: 2012
Date: 19 May
Mode: Online
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2012 · 19 May · Online · Mathematics · Trigonometric Ratios & Identities

Suppose θ and φ(≠ 0) are such that (θ+φ) , sec θ and (θ-φ) are in...

Suppose

\theta

and

\phi(\neq 0)

are such that

\sec (\theta+\phi)

, sec

\theta

and

\sec (\theta-\phi)

are in A.P. If

\cos \theta=k \cos

\left(\frac{\phi}{2}\right)

for some

, then

k is equal to

Options

A $\pm \sqrt{2}$
B $\pm 1$
C $\pm \frac{1}{\sqrt{2}}$
D $\pm 2$

Correct answer

A. $\pm \sqrt{2}$

Step-by-step solution

Since,

\sec (\theta-\phi), \sec \theta

and

\sec (\theta+\phi)

are in A.P.,

\begin{aligned}& \therefore 2 \sec \theta=\sec (\theta-\phi)+\sec (\theta+\phi)\\ & \Rightarrow \frac{2}{\cos \theta}=\frac{\cos (\theta+\phi)+\cos (\theta-\phi)}{\cos (\theta-\phi) \cos (\theta+\phi)} \end{aligned}

\begin{aligned}& \Rightarrow 2\left(\cos ^2 \theta-\sin ^2 \phi\right)=\cos \theta[2 \cos \theta \cos \phi\\ & \Rightarrow \cos ^2 \theta(1-\cos \phi)=\sin ^2 \phi=1-\cos ^2 \phi \\ & \Rightarrow \quad \cos ^2 \theta=1+\cos \phi=2 \cos ^2 \frac{\phi}{2}\\ & \therefore \quad \cos \theta=\pm \sqrt{2} \cos \frac{\phi}{2}\\ & \text{ But given } \cos \theta=\mathrm{k} \cos \frac{\phi}{2}\\ & \therefore \quad \mathrm{k}=\pm \sqrt{2} \end{aligned}

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