$1+2 \tan ^2 \theta+2 \sin \theta \sec ^2 \theta, 0^{\circ}<\theta<90^{\circ}$, is equal to :

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

1 + 2 tan²θ + 2 sinθ . sec²θ 

= 1 + 2 × \(\frac{sin²θ}{cos²θ}\)+ 2 × sinθ×  \(\frac{ 1 }{cos²θ}\)

= \(\frac{ cos²θ + 2 sin²θ + 2 sinθ }{cos²θ}\)

{ using , sin²θ + cos²θ = 1 }

 = \(\frac{ 1 +  sin²θ + 2 sinθ }{ 1 - sin²θ}\)

= \(\frac{ (1 + sinθ)²  }{ (1 - sinθ).(1 +sinθ) }\)

= \(\frac{ (1 + sinθ)  }{ (1 -sinθ) }\)