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Prove that:

(a) [tex] \frac{\cos \theta}{\operatorname{cosec} \theta + 1} + \frac{\cos \theta}{\operatorname{cosec} \theta - 1} = 2 \tan \theta [/tex]



Answer :

GinnyAnswer
To prove the given equation [tex]\(\frac{\cos \theta}{\operatorname{cosec} \theta + 1} + \frac{\cos \theta}{\operatorname{cosec} \theta - 1} = 2 \tan \theta\)[/tex], let's go through the steps systematically.

1. Express [tex]\(\operatorname{cosec} \theta\)[/tex]:
[tex]\[ \operatorname{cosec} \theta = \frac{1}{\sin \theta} \][/tex]

2. Substitute [tex]\(\operatorname{cosec} \theta\)[/tex] into the given equation:
[tex]\[ \frac{\cos \theta}{\frac{1}{\sin \theta} + 1} + \frac{\cos \theta}{\frac{1}{\sin \theta} - 1} \][/tex]

3. Simplify the denominators:
[tex]\[ \frac{\cos \theta}{\frac{1 + \sin \theta}{\sin \theta}} + \frac{\cos \theta}{\frac{1 - \sin \theta}{\sin \theta}} \][/tex]
Which simplifies to:
[tex]\[ \frac{\cos \theta \sin \theta}{1 + \sin \theta} + \frac{\cos \theta \sin \theta}{1 - \sin \theta} \][/tex]

4. Combine the fractions:
[tex]\[ \cos \theta \sin \theta \left(\frac{1}{1 + \sin \theta} + \frac{1}{1 - \sin \theta}\right) \][/tex]

5. Find common denominator:
[tex]\[ \cos \theta \sin \theta \left(\frac{1 - \sin \theta + 1 + \sin \theta}{(1 + \sin \theta)(1 - \sin \theta)}\right) \][/tex]
Notice that:
[tex]\[ (1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta \][/tex]

So:
[tex]\[ \cos \theta \sin \theta \left(\frac{2}{\cos^2 \theta}\right) \][/tex]

6. Simplify the expression:
[tex]\[ \frac{2 \cos \theta \sin \theta}{\cos^2 \theta} = 2 \left(\frac{\sin \theta}{\cos \theta}\right) = 2 \tan \theta \][/tex]

Therefore, we have shown that:

[tex]\[ \frac{\cos \theta}{\operatorname{cosec} \theta + 1} + \frac{\cos \theta}{\operatorname{cosec} \theta - 1} = 2 \tan \theta \][/tex]

This completes the proof.

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