cos (α-β)=1 and cos (α+β)=(1/e), where α, β ∈[-π, π]. Number of pairs (α, β) which satisfy both the equations is/are

Question

cos (α-β)=1 and cos (α+β)=(1/e), where α, β ∈[-π, π]. Number of pairs (α, β) which satisfy both the equations is/are (A) 0 (B) 1 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a7321bbb55fb3acb74adaa3106055c0b-.png" /> α-β=0,-2 π text or 2 π α-β=0 ⇒ α=β ⇒ cos 2 β=(1/e) This is true for ' 4 ' values of ' α ', ' β ' If α-β=-2 π ⇒ α=-π and β=π and cos (α+β)=1 ⇒ (No solution) similarly if α-β=2 π ⇒ α=π and β=-π again no solution results

Q. cos(α−β)=1 and cos(α+β)=e1​, where α,β∈[−π,π]. Number of pairs (α,β) which satisfy both the equations is/are

0

1

2

4

α−β=0,−2π or 2π α−β=0 ⇒α=β⇒cos2β=e1​ This is true for ' 4 ' values of ' α ', ' β ' If α−β=−2π⇒α=−π and β=π and cos(α+β)=1⇒ (No solution) similarly if α−β=2π⇒α=π and β=−π again no solution results

Q. $cos (α - β) = 1$ and $cos (α + β) = \frac{1}{e}$ , where $α, β \in [- π, π]$ . Number of pairs $(α, β)$ which satisfy both the equations is/are