Question

underot 1+sin^/1-sin^ = 1 + sin  cos 

Answer 1

Answer:

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Answer 2

Answer:

(secθ+tanθ)²

Step-by-step explanation:

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

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

= $\frac{(1+sin\theta)^2 }{(1-sin\theta) (1+sin\theta) }$

= $\frac{(1+sin\theta)^2 }{(1^2-sin^2\theta)}$

= $\frac{(1+sin\theta)^2 }{cos^2\theta}$

= $[\frac{1+sin\theta}{cos\theta}] ^2$

= (secθ+tanθ)²