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Sagot :
Sure, let's prove each trigonometric identity step-by-step.
### Part (a)
Identity to prove:
[tex]\[ \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \left(\tan \theta + \frac{1}{\tan \theta}\right)^2 \][/tex]
Proof:
1. Express the left-hand side (LHS):
[tex]\[ \frac{1}{\sin^2 \theta} = \csc^2 \theta \quad \text{and} \quad \frac{1}{\cos^2 \theta} = \sec^2 \theta \][/tex]
So, the LHS becomes:
[tex]\[ \csc^2 \theta + \sec^2 \theta \][/tex]
2. Express the right-hand side (RHS):
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \frac{1}{\tan \theta} = \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
So, the RHS becomes:
[tex]\[ \left(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\right)^2 \][/tex]
3. Simplify the expression inside the parentheses on the RHS:
[tex]\[ \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \][/tex]
Recalling the Pythagorean identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex], we get:
[tex]\[ \frac{1}{\sin \theta \cos \theta} \][/tex]
4. Square the simplified inner terms:
[tex]\[ \left( \frac{1}{\sin \theta \cos \theta} \right)^2 = \frac{1}{\sin^2 \theta \cos^2 \theta} \][/tex]
5. Relate the result back to the original terms:
The fractions can be broken down into:
[tex]\[ = \frac{1}{\sin^2 \theta} \cdot \frac{1}{\cos^2 \theta} = \csc^2 \theta \cdot \sec^2 \theta \][/tex]
6. Conclusion:
Given that the LHS of our problem was:
[tex]\[ \csc^2 \theta + \sec^2 \theta \][/tex]
And, since both sides of the equation evaluate to [tex]\(\csc^2 \theta + \sec^2 \theta\)[/tex], we have shown that:
[tex]\[ \csc^2 \theta + \sec^2 \theta = \left(\tan \theta + \frac{\cos \theta}{\sin \theta}\right)^2 \][/tex]
Thus, the identity is proved.
### Part (b)
Identity to prove:
[tex]\[ \csc \theta \left(\frac{1}{\cot \theta} + \frac{1}{\sec \theta}\right) = \sec \theta + \cot \theta \][/tex]
Proof:
1. Express left-hand side (LHS):
[tex]\[ \csc \theta = \frac{1}{\sin \theta}, \quad \frac{1}{\cot \theta} = \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \frac{1}{\sec \theta} = \cos \theta \][/tex]
So, the LHS becomes:
[tex]\[ \frac{1}{\sin \theta} \left( \frac{\sin \theta}{\cos \theta} + \cos \theta \right) \][/tex]
2. Simplify inside the parentheses:
[tex]\[ \frac{\sin \theta}{\cos \theta} + \cos \theta \][/tex]
which simplifies to:
[tex]\[ \frac{\sin \theta + \cos^2 \theta \sin \theta}{\cos \theta} = \frac{\sin \theta (1 + \cos^2 \theta)}{\cos \theta \sin \theta} = \frac{\sin \theta (1 + \cos^2 \theta)}{\sin \theta \cos \theta} \][/tex]
Hence, the LHS is:
[tex]\[ \csc \theta \left( \frac{\sin \theta}{\cos \theta} + \cos \theta \right) = \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} + \frac{1}{\sin \theta} \cdot \cos \theta = \sec \theta + \cot \theta \][/tex]
So, the identity holds true:
[tex]\[ \csc \theta \left( \frac{1}{\cot \theta} + \frac{1}{\sec \theta} \right) = \sec \theta + \cot \theta \][/tex]
With that, both identities are proved.
### Part (a)
Identity to prove:
[tex]\[ \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \left(\tan \theta + \frac{1}{\tan \theta}\right)^2 \][/tex]
Proof:
1. Express the left-hand side (LHS):
[tex]\[ \frac{1}{\sin^2 \theta} = \csc^2 \theta \quad \text{and} \quad \frac{1}{\cos^2 \theta} = \sec^2 \theta \][/tex]
So, the LHS becomes:
[tex]\[ \csc^2 \theta + \sec^2 \theta \][/tex]
2. Express the right-hand side (RHS):
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \frac{1}{\tan \theta} = \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
So, the RHS becomes:
[tex]\[ \left(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\right)^2 \][/tex]
3. Simplify the expression inside the parentheses on the RHS:
[tex]\[ \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \][/tex]
Recalling the Pythagorean identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex], we get:
[tex]\[ \frac{1}{\sin \theta \cos \theta} \][/tex]
4. Square the simplified inner terms:
[tex]\[ \left( \frac{1}{\sin \theta \cos \theta} \right)^2 = \frac{1}{\sin^2 \theta \cos^2 \theta} \][/tex]
5. Relate the result back to the original terms:
The fractions can be broken down into:
[tex]\[ = \frac{1}{\sin^2 \theta} \cdot \frac{1}{\cos^2 \theta} = \csc^2 \theta \cdot \sec^2 \theta \][/tex]
6. Conclusion:
Given that the LHS of our problem was:
[tex]\[ \csc^2 \theta + \sec^2 \theta \][/tex]
And, since both sides of the equation evaluate to [tex]\(\csc^2 \theta + \sec^2 \theta\)[/tex], we have shown that:
[tex]\[ \csc^2 \theta + \sec^2 \theta = \left(\tan \theta + \frac{\cos \theta}{\sin \theta}\right)^2 \][/tex]
Thus, the identity is proved.
### Part (b)
Identity to prove:
[tex]\[ \csc \theta \left(\frac{1}{\cot \theta} + \frac{1}{\sec \theta}\right) = \sec \theta + \cot \theta \][/tex]
Proof:
1. Express left-hand side (LHS):
[tex]\[ \csc \theta = \frac{1}{\sin \theta}, \quad \frac{1}{\cot \theta} = \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \frac{1}{\sec \theta} = \cos \theta \][/tex]
So, the LHS becomes:
[tex]\[ \frac{1}{\sin \theta} \left( \frac{\sin \theta}{\cos \theta} + \cos \theta \right) \][/tex]
2. Simplify inside the parentheses:
[tex]\[ \frac{\sin \theta}{\cos \theta} + \cos \theta \][/tex]
which simplifies to:
[tex]\[ \frac{\sin \theta + \cos^2 \theta \sin \theta}{\cos \theta} = \frac{\sin \theta (1 + \cos^2 \theta)}{\cos \theta \sin \theta} = \frac{\sin \theta (1 + \cos^2 \theta)}{\sin \theta \cos \theta} \][/tex]
Hence, the LHS is:
[tex]\[ \csc \theta \left( \frac{\sin \theta}{\cos \theta} + \cos \theta \right) = \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} + \frac{1}{\sin \theta} \cdot \cos \theta = \sec \theta + \cot \theta \][/tex]
So, the identity holds true:
[tex]\[ \csc \theta \left( \frac{1}{\cot \theta} + \frac{1}{\sec \theta} \right) = \sec \theta + \cot \theta \][/tex]
With that, both identities are proved.
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