How do you simplify the expression \[{{\csc }^{2}}\theta -{{\cot }^{2}}\theta \]?
Answer
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Hint: From the question given, we have been asked to simplify \[{{\csc }^{2}}\theta -{{\cot }^{2}}\theta \]. To solve the given, we have to use the basic formulae of trigonometry relating sine and cosine functions with cot and cosec functions. After using the basic formulae of trigonometry to the given question, we have to simplify further to get the final accurate and exact answer.
Complete step-by-step solution:
From the question, we have been given that,
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta \]
From the basic formulae of trigonometry, we already know that,
\[\Rightarrow \csc \theta =\dfrac{1}{\sin \theta }\]
And
\[\Rightarrow \cot \theta =\dfrac{\cos \theta }{\sin \theta }\]
Now, we have to substitute the above formula in the given question.
By simplifying the above obtained trigonometric expression further, we get
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =\dfrac{1}{{{\sin }^{2}}\theta }-\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]
By taking denominator common we get,
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =\dfrac{1-{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]
From the general identities of trigonometry, we already know that
\[\Rightarrow \text{ }\]\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
Therefore,
\[\Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \]
Now, we have to substitute the value of the above identity in the above trigonometric expression to get the final answer.
By substituting the value of the above identity in the above trigonometric expression, we get
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }\]
Therefore,
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1\]
Hence, the given question is simplified by using the basic formulae of trigonometry and general identity of trigonometry.
Note: Students should be well aware of the basic formulae of trigonometry and also be well aware of the general identities of the trigonometry. We have been given an identity in the question, so we must get the final answer as 1 no matter what. Students should be very careful while doing the calculation part by performing the correct substitutions for functions.
Complete step-by-step solution:
From the question, we have been given that,
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta \]
From the basic formulae of trigonometry, we already know that,
\[\Rightarrow \csc \theta =\dfrac{1}{\sin \theta }\]
And
\[\Rightarrow \cot \theta =\dfrac{\cos \theta }{\sin \theta }\]
Now, we have to substitute the above formula in the given question.
By simplifying the above obtained trigonometric expression further, we get
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =\dfrac{1}{{{\sin }^{2}}\theta }-\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]
By taking denominator common we get,
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =\dfrac{1-{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]
From the general identities of trigonometry, we already know that
\[\Rightarrow \text{ }\]\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
Therefore,
\[\Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \]
Now, we have to substitute the value of the above identity in the above trigonometric expression to get the final answer.
By substituting the value of the above identity in the above trigonometric expression, we get
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }\]
Therefore,
\[\Rightarrow {{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1\]
Hence, the given question is simplified by using the basic formulae of trigonometry and general identity of trigonometry.
Note: Students should be well aware of the basic formulae of trigonometry and also be well aware of the general identities of the trigonometry. We have been given an identity in the question, so we must get the final answer as 1 no matter what. Students should be very careful while doing the calculation part by performing the correct substitutions for functions.
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