Question

Consider the equation tan(270° + 0)= 1 - sin∅/cos∅ where 0° <∅ < 360°

Answer 1

$\\ \sf\longmapsto tan(270+\theta)=1-\dfrac{sin\theta}{cos\theta}$

$\\ \sf\longmapsto \dfrac{sin(270+\theta)}{cos(270+\theta)}=1-\dfrac{sin\theta}{cos\theta}$

$\\ \sf\longmapsto \dfrac{-cos\theta}{sin\theta}=1-\dfrac{sin\theta}{cos\theta}$

$\\ \sf\longmapsto\dfrac{sin\theta}{cos\theta} \dfrac{-cos\theta}{sin\theta}=1$

$\\ \sf\longmapsto \dfrac{sin^2\theta-cos^2\theta}{-cos\theta sin\theta}=1$

$\\ \sf\longmapsto \dfrac{-2cos\theta}{-cos\theta sin\theta}$

$\\ \sf\longmapsto \dfrac{2}{sin\theta}=1$

$\\ \sf\longmapsto 2cosec\theta=1$

$\\ \sf\longmapsto cosec\theta=\dfrac{1}{2}$

$\\ \sf\longmapsto \theta=cosec^{-1}\left(\dfrac{1}{2}\right)$

• theta doesn't exist