From science to arts, IDNLearn.com has the answers to all your questions. Our platform provides prompt, accurate answers from experts ready to assist you with any question you may have.
Sagot :
Let's examine the given equation and verify whether it holds true:
[tex]\[ \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A. \][/tex]
### Step 1: Simplify the left-hand side (LHS) of the equation.
Let [tex]\( x = \cos A + \sin A \)[/tex]. Then we can rewrite the LHS as:
[tex]\[ \frac{1}{x - 1} + \frac{1}{x + 1}. \][/tex]
### Step 2: Combine the fractions on the LHS.
We combine the fractions by obtaining a common denominator:
[tex]\[ \frac{1}{x - 1} + \frac{1}{x + 1} = \frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)} = \frac{x + 1 + x - 1}{(x - 1)(x + 1)} = \frac{2x}{x^2 - 1}. \][/tex]
Recall that [tex]\( x = \cos A + \sin A \)[/tex], so substituting [tex]\( x \)[/tex] back into the fraction:
[tex]\[ \frac{2(\cos A + \sin A)}{(\cos A + \sin A)^2 - 1}. \][/tex]
### Step 3: Simplify the expression.
[tex]\[ (\cos A + \sin A)^2 = \cos^2 A + \sin^2 A + 2 \cos A \sin A = 1 + 2 \cos A \sin A. \][/tex]
Then the denominator becomes:
[tex]\[ (\cos A + \sin A)^2 - 1 = (1 + 2 \cos A \sin A) - 1 = 2 \cos A \sin A. \][/tex]
Therefore, the LHS simplifies to:
[tex]\[ \frac{2(\cos A + \sin A)}{2 \cos A \sin A} = \frac{\cos A + \sin A}{\cos A \sin A}. \][/tex]
We can split this fraction into two separate terms:
[tex]\[ \frac{\cos A + \sin A}{\cos A \sin A} = \frac{\cos A}{\cos A \sin A} + \frac{\sin A}{\cos A \sin A} = \frac{1}{\sin A} + \frac{1}{\cos A} = \csc A + \sec A. \][/tex]
### Step 4: Compare the simplified LHS with the right-hand side (RHS).
So, we have:
[tex]\[ \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \csc A + \sec A. \][/tex]
Thus, we see that the LHS simplifies directly to the RHS:
[tex]\[ \sec A + \csc A = \sec A + \csc A. \][/tex]
Hence, we have shown that the given equation holds true.
[tex]\[ \boxed{\frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A}. \][/tex]
[tex]\[ \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A. \][/tex]
### Step 1: Simplify the left-hand side (LHS) of the equation.
Let [tex]\( x = \cos A + \sin A \)[/tex]. Then we can rewrite the LHS as:
[tex]\[ \frac{1}{x - 1} + \frac{1}{x + 1}. \][/tex]
### Step 2: Combine the fractions on the LHS.
We combine the fractions by obtaining a common denominator:
[tex]\[ \frac{1}{x - 1} + \frac{1}{x + 1} = \frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)} = \frac{x + 1 + x - 1}{(x - 1)(x + 1)} = \frac{2x}{x^2 - 1}. \][/tex]
Recall that [tex]\( x = \cos A + \sin A \)[/tex], so substituting [tex]\( x \)[/tex] back into the fraction:
[tex]\[ \frac{2(\cos A + \sin A)}{(\cos A + \sin A)^2 - 1}. \][/tex]
### Step 3: Simplify the expression.
[tex]\[ (\cos A + \sin A)^2 = \cos^2 A + \sin^2 A + 2 \cos A \sin A = 1 + 2 \cos A \sin A. \][/tex]
Then the denominator becomes:
[tex]\[ (\cos A + \sin A)^2 - 1 = (1 + 2 \cos A \sin A) - 1 = 2 \cos A \sin A. \][/tex]
Therefore, the LHS simplifies to:
[tex]\[ \frac{2(\cos A + \sin A)}{2 \cos A \sin A} = \frac{\cos A + \sin A}{\cos A \sin A}. \][/tex]
We can split this fraction into two separate terms:
[tex]\[ \frac{\cos A + \sin A}{\cos A \sin A} = \frac{\cos A}{\cos A \sin A} + \frac{\sin A}{\cos A \sin A} = \frac{1}{\sin A} + \frac{1}{\cos A} = \csc A + \sec A. \][/tex]
### Step 4: Compare the simplified LHS with the right-hand side (RHS).
So, we have:
[tex]\[ \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \csc A + \sec A. \][/tex]
Thus, we see that the LHS simplifies directly to the RHS:
[tex]\[ \sec A + \csc A = \sec A + \csc A. \][/tex]
Hence, we have shown that the given equation holds true.
[tex]\[ \boxed{\frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A}. \][/tex]
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.