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.
Similar Questions
What is the value of sin 26° + sin 212° + sin 218° + … + sin 284° + sin 290°?
Consider the following statements:
1. (sec 2θ - 1) (1 - cosec 2θ) = 1
2. sin θ (1 + cos θ) -1 + (1 + cos θ) (sin θ) -1 = 2 cosec θ
Which of the above is/are correct?
Important Questions from Trigonometric Identities
If cos 2θ = sin θ and θ lies between 0 and 90°, then θ will be:
\(\frac{3 - 4\sin^2 \theta}{\cos^2 \theta} + 2\tan^2 \theta\) can be simplified as:
Simplify \( \left(\frac{1}{\sin^2 A} - 1\right) \), where \( 0 < A \leq 90^\circ \).
If \( \tan \theta = \frac{8}{15} \), then the value of \( \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \) is:
If the volume of a cuboid is \(3x^2 - 27\), then its possible dimensions are:
