Prove that $\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$.
Explanation
We can simplify the given expression by multiplying the numerator and denominator by (1 - sin θ). This gives us: \\frac{1+\\sec \\theta-\\tan \\theta}{1+\\sec \\theta+\\tan \\theta}=\\frac{(1+\\sec \\theta-\\tan \\theta)(1-\\sin \\theta)}{(1+\\sec \\theta+\\tan \\theta)(1-\\sin \\theta)}=\\frac{1-\\sin \\theta+\\sec \\theta-\\sec \\theta \\sin \\theta-\\tan \\theta+\\tan \\theta \\sin \\theta}{1-\\sin \\theta+\\sec \\theta \\sin \\theta+\\sec \\theta+\\tan \\theta \\sin \\theta-\\tan \\theta}=\\frac{1-\\sin \\theta}{\\cos \\theta}
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