Problem 179
Question
$$ \tan \left(45^{\circ}+\frac{A}{2}\right)=\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A $$
Step-by-Step Solution
Verified Answer
To solve the problem, it is necessary to manipulate the given equation in order to express \(\tan \frac{A}{2}\) in terms of A.
1Step 1: Expand and Simplify the Left-Hand Side
Let's rewrite \(\tan \left(45^{\circ}+\frac{A}{2}\right)\) using the formula for tan of sum, which states that \(\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\). So, \(\tan \left(45^{\circ}+\frac{A}{2}\right) = \frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}}\). Now, the left-hand side of the equation is \(\frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}}\).
2Step 2: Simplify the Right-Hand Side
It is also advantageous to represent \(\sec A\) and \(\tan A\) in terms of \(\sin A\) and \(\cos A\). We know that \(\sec A = \frac{1}{\cos A}\) and \(\tan A= \frac{\sin A}{\cos A}\). So, the right-hand side becomes \( \frac{1 + \sin A}{\cos A}\)
3Step 3: Equate and Simplify
Now we can equate the simplified right and left sides: \(\frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}} = \frac{1 + \sin A}{\cos A}\). To make this equation easier to solve, let \(\tan \frac{A}{2} = x\) which simplifies the equation to: \(\frac{1 + x}{1 - x} = \frac{1 + \sin A}{\cos A}\)
4Step 4: Solve for x in terms of A
From the equation \(\frac{1 + x}{1 - x} = \frac{1 + \sin A}{\cos A}\), it is possible to isolate x ( \(\tan \frac{A}{2}\) ) and express it in terms of A. This expression is the main result of the problem
Key Concepts
Tangent FormulaSecant and Tangent RelationshipHalf-Angle Identity
Tangent Formula
The tangent formula is a fundamental trigonometric identity used to calculate the tangent of the sum of two angles. When you encounter \( \tan(A + B) \), it can be expanded using the formula:
This expansion helps you translate a trigonometric expression into a form that is easier to work with, often paving the way for further simplifications or transformations in the exercise.
- \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
This expansion helps you translate a trigonometric expression into a form that is easier to work with, often paving the way for further simplifications or transformations in the exercise.
Secant and Tangent Relationship
The secant and tangent functions are inherently related through the identities derived from sine and cosine. To clarify this relationship, recall the definitions:
In the step-by-step solution, the given expression \( \sec A + \tan A \) is simplified as \( \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} \).
Such transformations are essential when comparing or equating different sides of a trigonometrical equation, allowing you to manipulate and solve it more easily by simplifying complexity into a more manageable form.
- \( \sec A = \frac{1}{\cos A} \)
- \( \tan A = \frac{\sin A}{\cos A} \)
In the step-by-step solution, the given expression \( \sec A + \tan A \) is simplified as \( \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} \).
Such transformations are essential when comparing or equating different sides of a trigonometrical equation, allowing you to manipulate and solve it more easily by simplifying complexity into a more manageable form.
Half-Angle Identity
The half-angle identities are crucial in trigonometry for converting expressions involving angles into more workable forms. The identity for tangent of a half-angle states:
By defining \( \tan \frac{A}{2} \) as \( x \), the solution further simplifies the relationship between different trigonometric expressions on either side of an equation.
Such identities enable the transformation of difficult trigonometric computations into more solvable forms by reducing angles, thus making analysis and solutions more accessible.
- \( \tan \frac{A}{2} = \frac{1 - \cos A}{\sin A} \)
By defining \( \tan \frac{A}{2} \) as \( x \), the solution further simplifies the relationship between different trigonometric expressions on either side of an equation.
Such identities enable the transformation of difficult trigonometric computations into more solvable forms by reducing angles, thus making analysis and solutions more accessible.
Other exercises in this chapter
Problem 177
$$ (\cos \alpha+\cos \beta)^{2}+(\sin \alpha+\sin \beta)^{2}=4 \cos ^{2} \frac{\alpha-\beta}{2} $$
View solution Problem 178
$$ (\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}=4 \sin ^{2} \frac{\alpha-\beta}{2} $$
View solution Problem 180
$$ \left(1+\tan \frac{\alpha}{2}-\sec \frac{\alpha}{2}\right)\left(1+\tan \frac{\alpha}{2}+\sec \frac{\alpha}{2}\right)=\sin \alpha \sec ^{2} \frac{\alpha}{2} $
View solution Problem 181
$$ \frac{1-\cos A+\cos B-\cos (A+B)}{1+\cos A-\cos B-\cos (A+B)}=\tan \frac{A}{2} \cot \frac{B}{2} $$
View solution