Question

1 + sec theta - tan theta/1 + sec theta + tan theta = 1 + sin theta/ cos theta

Answer 1

I think you did mistake in typing. question should be --->

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

$LHS=\frac{1+sec\theta-tan\theta}{1+sec\theta+tan\theta}\\\\=\frac{sec^2\theta-tan^2\theta+sec\theta-tan\theta}{1+sec\theta+tan\theta}\\\\\textbf{we know},sec^2x-tan^2x=1\\\bf{so},\textbf{we can replace},sec^2\theta-tan^2\theta,\textbf{in place of},1\\\\=\frac{(sec\theta-tan\theta)(sec\theta+tan\theta)+(sec\theta-tan\theta)}{1+sec\theta+tan\theta}\\\\=\frac{(sec\theta-tan\theta)(1+sec\theta+tan\theta)}{1+sec\theta+tan\theta}\\\\=sec\theta-tan\theta\\\\=\frac{1}{cos\theta}-\frac{sin\theta}{cos\theta}\\\\=\frac{1-sin\theta}{cos\theta}=RHS$

Answer 2

Answer:

We have to find the value of the expression:

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

We will rationalize our denominator:

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

since,

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

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

Hence, the value of the expression is:

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