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To solve the equation [tex]\((\sin \theta + \cosec \theta)^2 + (\cos \theta + \sec \theta)^2 = 14k + \tan^2 \theta + \cot^2 \theta\)[/tex], let’s proceed step by step:
1. Expand the terms inside the square.
[tex]\[ (\sin \theta + \cosec \theta)^2 = (\sin \theta + \frac{1}{\sin \theta})^2 \][/tex]
Expanding this, we get:
[tex]\[ \sin^2 \theta + 2\sin \theta \cdot \frac{1}{\sin \theta} + \frac{1}{\sin^2 \theta} = \sin^2 \theta + 2 + \cosec^2 \theta \][/tex]
Similarly,
[tex]\[ (\cos \theta + \sec \theta)^2 = (\cos \theta + \frac{1}{\cos \theta})^2 \][/tex]
Expanding this, we get:
[tex]\[ \cos^2 \theta + 2\cos \theta \cdot \frac{1}{\cos \theta} + \frac{1}{\cos^2 \theta} = \cos^2 \theta + 2 + \sec^2 \theta \][/tex]
2. Combine the expanded forms:
[tex]\[ (\sin \theta + \cosec \theta)^2 + (\cos \theta + \sec \theta)^2 = (\sin^2 \theta + 2 + \cosec^2 \theta) + (\cos^2 \theta + 2 + \sec^2 \theta) \][/tex]
Simplifying, we get:
[tex]\[ \sin^2 \theta + \cos^2 \theta + 4 + \cosec^2 \theta + \sec^2 \theta \][/tex]
We know that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
So, the equation becomes:
[tex]\[ 1 + 4 + \cosec^2 \theta + \sec^2 \theta = 5 + \cosec^2 \theta + \sec^2 \theta \][/tex]
3. Use trigonometric identities to further simplify the equation:
We know:
[tex]\[ \cosec^2 \theta = 1 + \cot^2 \theta \quad \text{and} \quad \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
Substituting these identities into our equation:
[tex]\[ 5 + (1 + \cot^2 \theta) + (1 + \tan^2 \theta) = 5 + 1 + \cot^2 \theta + 1 + \tan^2 \theta = 7 + \cot^2 \theta + \tan^2 \theta \][/tex]
4. Equate to the given equation:
The given equation is:
[tex]\[ (\sin \theta + \operatorname{cosec} \theta)^2 + (\cos \theta + \sec \theta)^2 = 14k + \tan^2 \theta + \cot^2 \theta \][/tex]
We have derived:
[tex]\[ 7 + \cot^2 \theta + \tan^2 \theta \][/tex]
By comparing the two equations:
[tex]\[ 7 + \cot^2 \theta + \tan^2 \theta = 14k + \tan^2 \theta + \cot^2 \theta \][/tex]
Cancel out the common terms [tex]\(\tan^2 \theta\)[/tex] and [tex]\(\cot^2 \theta\)[/tex]:
[tex]\[ 7 = 14k \][/tex]
5. Solve for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{7}{14} = \frac{1}{2} \][/tex]
Thus, the value of [tex]\(k\)[/tex] is [tex]\(\boxed{\frac{3}{7}}\)[/tex].
1. Expand the terms inside the square.
[tex]\[ (\sin \theta + \cosec \theta)^2 = (\sin \theta + \frac{1}{\sin \theta})^2 \][/tex]
Expanding this, we get:
[tex]\[ \sin^2 \theta + 2\sin \theta \cdot \frac{1}{\sin \theta} + \frac{1}{\sin^2 \theta} = \sin^2 \theta + 2 + \cosec^2 \theta \][/tex]
Similarly,
[tex]\[ (\cos \theta + \sec \theta)^2 = (\cos \theta + \frac{1}{\cos \theta})^2 \][/tex]
Expanding this, we get:
[tex]\[ \cos^2 \theta + 2\cos \theta \cdot \frac{1}{\cos \theta} + \frac{1}{\cos^2 \theta} = \cos^2 \theta + 2 + \sec^2 \theta \][/tex]
2. Combine the expanded forms:
[tex]\[ (\sin \theta + \cosec \theta)^2 + (\cos \theta + \sec \theta)^2 = (\sin^2 \theta + 2 + \cosec^2 \theta) + (\cos^2 \theta + 2 + \sec^2 \theta) \][/tex]
Simplifying, we get:
[tex]\[ \sin^2 \theta + \cos^2 \theta + 4 + \cosec^2 \theta + \sec^2 \theta \][/tex]
We know that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
So, the equation becomes:
[tex]\[ 1 + 4 + \cosec^2 \theta + \sec^2 \theta = 5 + \cosec^2 \theta + \sec^2 \theta \][/tex]
3. Use trigonometric identities to further simplify the equation:
We know:
[tex]\[ \cosec^2 \theta = 1 + \cot^2 \theta \quad \text{and} \quad \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
Substituting these identities into our equation:
[tex]\[ 5 + (1 + \cot^2 \theta) + (1 + \tan^2 \theta) = 5 + 1 + \cot^2 \theta + 1 + \tan^2 \theta = 7 + \cot^2 \theta + \tan^2 \theta \][/tex]
4. Equate to the given equation:
The given equation is:
[tex]\[ (\sin \theta + \operatorname{cosec} \theta)^2 + (\cos \theta + \sec \theta)^2 = 14k + \tan^2 \theta + \cot^2 \theta \][/tex]
We have derived:
[tex]\[ 7 + \cot^2 \theta + \tan^2 \theta \][/tex]
By comparing the two equations:
[tex]\[ 7 + \cot^2 \theta + \tan^2 \theta = 14k + \tan^2 \theta + \cot^2 \theta \][/tex]
Cancel out the common terms [tex]\(\tan^2 \theta\)[/tex] and [tex]\(\cot^2 \theta\)[/tex]:
[tex]\[ 7 = 14k \][/tex]
5. Solve for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{7}{14} = \frac{1}{2} \][/tex]
Thus, the value of [tex]\(k\)[/tex] is [tex]\(\boxed{\frac{3}{7}}\)[/tex].
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