If tan -1( sin 2 θ+2 sin θ+2)+ cot -1(4 sec 2 φ+1)=(π/2) has solution for some θ and φ then

Question

If tan -1( sin 2 θ+2 sin θ+2)+ cot -1(4 sec 2 φ+1)=(π/2) has solution for some θ and φ then (A)  sin θ=-1 (B)  sin θ=1 (C)  cos φ=-1 (D)  cos φ=1

Tardigrade Admin · Accepted Answer

For given equation to hold true sin 2 θ+2 sin θ+2=4 sec 2 φ+1 text L.H.S. =( sin θ+1)2+1 ∴ 1 ≤( sin θ+1)2+1 ≤ 5 text and RHS ≥ 5 text as sec 2 φ ≥ 1 ∴ text LHS = text RHS =5 ∴ sin θ=1 text and sec 2 φ=1 ⇒ cos φ= ± 1

Q. If $tan^{- 1} (sin^{2} θ + 2 sin θ + 2) + cot^{- 1} (4^{s e c^{2} ϕ} + 1) = \frac{π}{2}$ has solution for some $θ$ and $ϕ$ then

$sin θ = - 1$

$sin θ = 1$

$cos ϕ = - 1$

$cos ϕ = 1$

For given equation to hold true
$sin^{2} θ + 2 sin θ + 2 = 4^{s e c^{2} ϕ + 1}$
$L.H.S. = (sin θ + 1)^{2} + 1$
$∴ 1 \leq (sin θ + 1)^{2} + 1 \leq 5 and RHS \geq 5 as sec^{2} ϕ \geq 1$
$∴ LHS = RHS = 5$
$∴ sin θ = 1 and sec^{2} ϕ = 1 \Rightarrow cos ϕ = \pm 1$

Q. If tan−1(sin2θ+2sinθ+2)+cot−1(4sec2ϕ+1)=2π​ has solution for some θ and ϕ then

sinθ=−1

sinθ=1

cosϕ=−1

cosϕ=1