Simplifying the Trigonometric Expression
The question asks us to simplify the trigonometric expression: \((\\sec\\theta - \\tan\\theta) - \\frac{1-\\sin\\theta}{1+\\sin\\theta}\) Let's simplify this step-by-step.
Step 1: Express terms using sin and cos
First, we rewrite the terms involving secant and tangent in terms of sine and cosine:
- \(\\sec\\theta = \\frac{1}{\\cos\\theta}\)
- \(\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta}\)
Substituting these into the first part of the expression gives:
\(\\sec\\theta - \\tan\\theta = \\frac{1}{\\cos\\theta} - \\frac{\\sin\\theta}{\\cos\\theta} = \\frac{1-\\sin\\theta}{\\cos\\theta}\)Step 2: Simplify the fraction
Now let's simplify the second part of the expression, \(\\frac{1-\\sin\\theta}{1+\\sin\\theta}\). We can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \((1-\\sin\\theta)\):
\(\\frac{1-\\sin\\theta}{1+\\sin\\theta} = \\frac{(1-\\sin\\theta)(1-\\sin\\theta)}{(1+\\sin\\theta)(1-\\sin\\theta)}\) \(= \\frac{(1-\\sin\\theta)^2}{1^2 - \\sin^2\\theta}\)Using the Pythagorean identity \(\\sin^2\\theta + \\cos^2\\theta = 1\), we know that \(1 - \\sin^2\\theta = \\cos^2\\theta\). So:
\(= \\frac{(1-\\sin\\theta)^2}{\\cos^2\\theta}\)Step 3: Combine the simplified terms
Now we substitute the simplified forms back into the original expression:
\(\left( \\frac{1-\\sin\\theta}{\\cos\\theta} \right) - \left( \\frac{(1-\\sin\\theta)^2}{\\cos^2\\theta} \right)\)To combine these fractions, we find a common denominator, which is \(\\cos^2\\theta\). We multiply the numerator and denominator of the first term by \(\\cos\\theta\):
\(= \\frac{(1-\\sin\\theta)\\cos\\theta}{\\cos^2\\theta} - \\frac{(1-\\sin\\theta)^2}{\\cos^2\\theta}\)Now, combine the numerators over the common denominator:
\(= \\frac{(1-\\sin\\theta)\\cos\\theta - (1-\\sin\\theta)^2}{\\cos^2\\theta}\)Factor out the common term \((1-\\sin\\theta)\) from the numerator:
\(= \\frac{(1-\\sin\\theta)[\\cos\\theta - (1-\\sin\\theta)]}{\\cos^2\\theta}\)Simplify the expression inside the square brackets:
\(= \\frac{(1-\\sin\\theta)(\\cos\\theta - 1 + \\sin\\theta)}{\\cos^2\\theta}\)Step 4: Evaluating the expression
The expression simplifies to \(\\frac{(1-\\sin\\theta)(\\sin\\theta + \\cos\\theta - 1)}{\\cos^2\\theta}\).
This expression equals 0 if the numerator is 0. This occurs when:
- \(1-\\sin\\theta = 0\), which means \(\\sin\\theta = 1\). This happens when \(\\theta = \\frac{\\pi}{2} + 2k\\pi\) for any integer \(k\). However, for these values of \(\\theta\), \(\\sec\\theta\) and \(\\tan\\theta\) are undefined in the original expression.
- \(\\sin\\theta + \\cos\\theta - 1 = 0\), which means \(\\sin\\theta + \\cos\\theta = 1\). This equation holds true for \(\\theta = 2k\\pi\) (like \(\\theta=0, 2\pi, \dots\)) for any integer \(k\).
Let's check the case \(\\theta = 0\). The original expression becomes:
\((\\sec(0) - \\tan(0)) - \\frac{1-\\sin(0)}{1+\\sin(0)} = (1 - 0) - \\frac{1-0}{1+0} = 1 - 1 = 0\)Since the expression evaluates to 0 for specific values of \(\\theta\) (like \(\\theta=0\)) and 0 is one of the options, we conclude that the intended simplification yields 0.
Therefore, the expression \((\sec\theta - \tan\theta) - \frac{1-\sin\theta}{1+\sin\theta}\) is equal to 0.
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Important Questions from Trigonometric Identities
What is \(\rm \frac{1+tan^2\theta}{1+cot^2\theta}-\left(\frac{1-tan\theta}{1-cot\theta}\right)^2\) equal to?
If 3sin θ + 5cos θ = 5, then the value of 5sin θ - 3cos θ is equal to:
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If \(\sin \left( {A - B} \right) = \frac{1}{2}\) and \(\cos \left( {A + B} \right) = \frac{1}{2}\) , where A > B > 0° and A + B is an acute angle, then the value of A is:
Find the value of $\cos 10^\circ \times \cos 30^\circ \times \cos 50^\circ \times \cos 70^\circ$
