Question

prove that-
ta theta + sin theta/ tan theta - sin theta = sec theta + 1/ sec theta - 1

Answer 1

Given that
$\frac{tan \theta + sin \theta}{tan \theta - sin \theta} = \frac{sec \theta + 1}{sec \theta - 1}$

Consider LHS

$\frac{tan \theta + sin \theta}{tan \theta - sin \theta}$

$\frac{Sin\theta( \frac{1}{cos\theta} + 1)}{Sin\theta( \frac{1}{cos\theta} - 1)}}$

$\frac{( \frac{1+cos\theta}{cos\theta})}{(\frac{1-cos\theta}{cos\theta})}}$

$\frac{(1+cos\theta)}{({1-cos\theta)}}$

$\frac{1+ \frac{1}{sec\theta} }{1- \frac{1}{sec\theta} }$

$\frac{sec\theta + 1}{sec\theta - 1}$

Answer 2

(tan θ+sinθ)/(tanθ-sinθ)

⇒{(sinθ/cosθ)+sinθ}/{sinθ/cosθ-sinθ}
⇒sinθ(secθ+1)/sinθ(secθ-1)
⇒(secθ+1)/(secθ-1)