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Sagot :
Sure, let's solve the given trigonometric expression step by step:
Given expression:
[tex]\[ \frac{\operatorname{cosec} \theta}{\sin \theta} - \frac{\cot \theta}{\tan \theta} = 1 \][/tex]
First, recall the trigonometric identities:
[tex]\[ \operatorname{cosec} \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Now, substitute these identities into the expression:
For [tex]\(\frac{\operatorname{cosec} \theta}{\sin \theta}\)[/tex]:
[tex]\[ \frac{\operatorname{cosec} \theta}{\sin \theta} = \frac{\frac{1}{\sin \theta}}{\sin \theta} = \frac{1}{\sin \theta \cdot \sin \theta} = \frac{1}{\sin^2 \theta} \][/tex]
For [tex]\(\frac{\cot \theta}{\tan \theta}\)[/tex]:
[tex]\[ \frac{\cot \theta}{\tan \theta} = \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
So the original expression now becomes:
[tex]\[ \frac{1}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Combine the terms over a common denominator:
[tex]\[ \frac{1 - \cos^2 \theta}{\sin^2 \theta} \][/tex]
Recall the Pythagorean identity:
[tex]\[ 1 - \cos^2 \theta = \sin^2 \theta \][/tex]
Substitute this identity back into the expression:
[tex]\[ \frac{\sin^2 \theta}{\sin^2 \theta} = 1 \][/tex]
Thus, we have shown step-by-step that the given expression simplifies to:
[tex]\[ 1 \][/tex]
Hence, it is true that:
[tex]\[ \frac{\operatorname{cosec} \theta}{\sin \theta} - \frac{\cot \theta}{\tan \theta} = 1 \][/tex]
Given expression:
[tex]\[ \frac{\operatorname{cosec} \theta}{\sin \theta} - \frac{\cot \theta}{\tan \theta} = 1 \][/tex]
First, recall the trigonometric identities:
[tex]\[ \operatorname{cosec} \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Now, substitute these identities into the expression:
For [tex]\(\frac{\operatorname{cosec} \theta}{\sin \theta}\)[/tex]:
[tex]\[ \frac{\operatorname{cosec} \theta}{\sin \theta} = \frac{\frac{1}{\sin \theta}}{\sin \theta} = \frac{1}{\sin \theta \cdot \sin \theta} = \frac{1}{\sin^2 \theta} \][/tex]
For [tex]\(\frac{\cot \theta}{\tan \theta}\)[/tex]:
[tex]\[ \frac{\cot \theta}{\tan \theta} = \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
So the original expression now becomes:
[tex]\[ \frac{1}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Combine the terms over a common denominator:
[tex]\[ \frac{1 - \cos^2 \theta}{\sin^2 \theta} \][/tex]
Recall the Pythagorean identity:
[tex]\[ 1 - \cos^2 \theta = \sin^2 \theta \][/tex]
Substitute this identity back into the expression:
[tex]\[ \frac{\sin^2 \theta}{\sin^2 \theta} = 1 \][/tex]
Thus, we have shown step-by-step that the given expression simplifies to:
[tex]\[ 1 \][/tex]
Hence, it is true that:
[tex]\[ \frac{\operatorname{cosec} \theta}{\sin \theta} - \frac{\cot \theta}{\tan \theta} = 1 \][/tex]
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