Question

tantheta+sectheta-1/tantheta-sectheta+1=(1+sintheta)/costheta

Answer 1

Answer:

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

Using

$\boxed{\bf\:sec^2A-tan^2A=1}$

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

Using

$\boxed{\bf\:a^2-b^2=(a-b)(a+b)}$

$=\frac{tan\theta+sec\theta-(sec\theta-tan\theta)(sec\theta+tan\theta)}{tan\theta-sec\theta+1}$

$=\frac{(sec\theta+tan\theta)(1-(sec\theta-tan\theta))}{tan\theta-sec\theta+1}$

Rearranging terms, we get

$=\frac{(sec\theta+tan\theta)(tan\theta-sec\theta+1)}{tan\theta-sec\theta+1}$

$=sec\theta+tan\theta$

$=\frac{1}{cos\theta}+\frac{sin\theta}{cos\theta}$

$=\frac{1+sin\theta}{cos\theta}$

$\implies\boxed{\frac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=\frac{1+sin\theta}{cos\theta}}$