Question

tan theta + secant theta minus 1 by tan theta minus secant theta + 1 equal to 1 + sin theta by
cos theta with the statement

Answer 1

See the attachment for detail solution
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Answer 2

Answer:

We have to prove the following equality:

$\dfrac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\dfrac{1+\sin \theta}{\cos\theta}$

on solving the left hand side of the equality we get:

firstly we will rationalize our denominator to obtain the following expression as:

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

Since, we know that:

$\sec^2\theta-\tan^2\theta=1$

Hence,

we get:

$=\dfrac{2\tan\theta(\tan\theta+\sec\theta)}{2\tan\theta}\\\\=\tan\theta+\sec\theta\\\\=\dfrac{\sin\theta}{\cos\theta}+\dfrac{1}{\cos\theta}\\\\=\dfrac{1+\sin\theta}{\cos\theta}$

Hence, we get the result:

$\dfrac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\dfrac{1+\sin \theta}{\cos\theta}$