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Question: How do you simplify \[\dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \the...

How do you simplify cot⁡(θ)csc⁡(θ)−sin⁡(θ)\dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \theta \right)} ?

Explanation

Solution

From the question given, we have been asked to simplify cot⁡(θ)csc⁡(θ)−sin⁡(θ)\dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \theta \right)}.
To solve the given, we have to use the basic formulae of trigonometry like csc⁡=1sin⁡\csc =\dfrac{1}{\sin } and cot⁡=cos⁡sin⁡\cot =\dfrac{\cos }{\sin } to simplify. So, we will substitute these in the given expression and then take LCM to simplify the expression further. We will also make use of the identity sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 to obtain the final result.

Complete step by step solution:
From the question, we have been given that, ⇒cot⁡(θ)csc⁡(θ)−sin⁡(θ)\Rightarrow \dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \theta \right)}
From the basic formulae of trigonometry, we already know that,
⇒csc⁡θ=1sin⁡θ\Rightarrow \csc \theta =\dfrac{1}{\sin \theta }
And
⇒cot⁡θ=cos⁡θsin⁡θ\Rightarrow \cot \theta =\dfrac{\cos \theta }{\sin \theta }
Now, we have to substitute the above formula in the given question.
By substituting the above formula in the given question, we get
⇒cos⁡(θ)sin⁡(θ)1sin⁡(θ)−sin⁡(θ)\Rightarrow \dfrac{\dfrac{\cos \left( \theta \right)}{\sin \left( \theta \right)}}{\dfrac{1}{\sin \left( \theta \right)}-\sin \left( \theta \right)}
⇒cos⁡(θ)sin⁡(θ)1−sin⁡2(θ)sin⁡(θ)\Rightarrow \dfrac{\dfrac{\cos \left( \theta \right)}{\sin \left( \theta \right)}}{\dfrac{1-{{\sin }^{2}}\left( \theta \right)}{\sin \left( \theta \right)}}
Now, as we have been already discussed earlier, we have to simplify further to get the exact answer for the given question.
By simplifying the above obtained trigonometric expression further, we get
⇒cot⁡(θ)csc⁡(θ)−sin⁡(θ)=cos⁡(θ)1−sin⁡2(θ)\Rightarrow \dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \theta \right)}=\dfrac{\cos \left( \theta \right)}{1-{{\sin }^{2}}\left( \theta \right)}
From the general identities of trigonometry, we already know that
⇒sin⁡2θ+cos⁡2θ=1\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Therefore,
⇒1−sin⁡2(θ)=cos⁡2(θ)\Rightarrow 1-{{\sin }^{2}}\left( \theta \right)={{\cos }^{2}}\left( \theta \right)
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
⇒cot⁡(θ)csc⁡(θ)−sin⁡(θ)=cos⁡(θ)cos⁡2(θ)\Rightarrow \dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \theta \right)}=\dfrac{\cos \left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}
⇒cot⁡(θ)csc⁡(θ)−sin⁡(θ)=1cos⁡(θ)\Rightarrow \dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \theta \right)}=\dfrac{1}{\cos \left( \theta \right)}
We know that
⇒1cos⁡(θ)=sec⁡(θ)\Rightarrow \dfrac{1}{\cos \left( \theta \right)}=\sec \left( \theta \right)
Therefore,
⇒cot⁡(θ)csc⁡(θ)−sin⁡(θ)=sec⁡(θ)\Rightarrow \dfrac{\cot \left( \theta \right)}{\csc \left( \theta \right)-\sin \left( \theta \right)}=\sec \left( \theta \right)

Note: Students should be well aware of the basic formulae of trigonometry and also be well aware of the general identities of the trigonometry. Whenever we get questions like this, students should try to convert them in the terms of sin and cos to be able to simplify easily. Make sure students should substitute cot⁡=cos⁡sin⁡\cot =\dfrac{\cos }{\sin } and not as sin⁡cos⁡\dfrac{\sin }{\cos } it will change the solutions completely. Recollect the identities and formulas correctly before solving trigonometric questions.