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

Question

Suppose θ and φ(≠ 0) are such that sec(θ+φ), sec θ and sec(θ-φ) are in A.P. If cos θ=k cos((φ/2)) for some k, then k is equal to (A) ±√2 (B) ±1 (C) ±(1/√2) (D) ±2

Tardigrade Admin · Accepted Answer

Since, sec(θ-φ), secθ and ec(θ+φ) are in A.P., ∴ 2 secθ=sec(θ-φ)+sec(θ +φ ) ⇒ (1/cos θ)=(cos(θ+φ )+cos(θ +φ )/cos(θ -φ )cos(θ +φ )) ⇒ 2(cos2 θ-sin2 φ)=cos θ[2 cos θ cos φ ] ⇒ cos2 θ=1+cos φ=2 cos2 (φ/2) ∴ cos θ=±√2cos (φ/2) But given cosθ=k cos (φ/2) ∴ k=±√2

Q. 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

$\pm 2$

$\pm 1$

$\pm \frac{1}{2}$

$\pm 2$

Q. 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

±2​

±1

±2​1​

±2

Since, sec(θ−ϕ),secθ and ec(θ+ϕ) are in A.P., ∴2secθ=sec(θ−ϕ)+sec(θ+ϕ) ⇒cosθ1​=cos(θ−ϕ)cos(θ+ϕ)cos(θ+ϕ)+cos(θ+ϕ)​ ⇒2(cos2θ−sin2ϕ)=cosθ[2cosθcosϕ] ⇒cos2θ=1+cosϕ=2cos22ϕ​ ∴cosθ=±2​cos2ϕ​ But given cosθ=kcos2ϕ​ ∴k=±2​

Q. 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

$\pm 2$

$\pm 1$

$\pm \frac{1}{2}$

$\pm 2$