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Question: How do you prove\[\sin \theta + \cot \theta \cos \theta = \csc \theta \]?...

How do you provesin⁡θ+cot⁡θcos⁡θ=csc⁡θ\sin \theta + \cot \theta \cos \theta = \csc \theta ?

Explanation

Solution

The very first step to solve this problem, replace cos⁡2x{\cos ^2}xwith 1sin⁡θ\dfrac{1}{{\sin \theta }}. After this take the sin⁡θ\sin \theta on the LHS and take it to RHS. After this, solve and simplify the RHS. You will get a term 1−sin⁡2x1 - {\sin ^2}xon the RHS. Replace this term by cos⁡2x{\cos ^2}x. Now, we will rearrange the RHS in such a way that we will end up with the LHS.

Formulas used: Use the given below formulas to solve the problem

cot⁡θ=1tan⁡θ sin⁡θ=1csc⁡θ cos⁡θ=1sec⁡θ  \cot \theta = \dfrac{1}{{\tan \theta }} \\\ \sin \theta = \dfrac{1}{{\csc \theta }} \\\ \cos \theta = \dfrac{1}{{\sec \theta }} \\\

Complete step-by-step solution:
The given question we have issin⁡θ+cot⁡θcos⁡θ=csc⁡θ\sin \theta + \cot \theta \cos \theta = \csc \theta
Before heading on to solve the above problem. We need to keep in mind few basic trigonometric formulas which will help you in almost every problem that you will face in this chapter.
And those special formulas are:-

cot⁡θ=1tan⁡θ sin⁡θ=1csc⁡θ cos⁡θ=1sec⁡θ  \cot \theta = \dfrac{1}{{\tan \theta }} \\\ \sin \theta = \dfrac{1}{{\csc \theta }} \\\ \cos \theta = \dfrac{1}{{\sec \theta }} \\\

Now, beginning to solve the problem, we will replace csc⁡θ\csc \theta on the RHS to 1sin⁡θ\dfrac{1}{{\sin \theta }}
Doing this we will end up with
⇒sin⁡θ+cot⁡θcos⁡θ=1sinθ\Rightarrow \sin \theta + \cot \theta \cos \theta = \dfrac{1}{{sin\theta }}
Taking the sin⁡θ\sin \theta to the RHS,
⇒cot⁡θcos⁡θ=1sin⁡θ−sin⁡θ ⇒cot⁡θcos⁡θ=1−sin⁡2θsin⁡θ  \Rightarrow \cot \theta \cos \theta = \dfrac{1}{{\sin \theta }} - \sin \theta \\\ \Rightarrow \cot \theta \cos \theta = \dfrac{{1 - {{\sin }^2}\theta }}{{\sin \theta }} \\\
Now, to solve this question we will take the RHS of the equation and solve it in such a manner that we will get the LHS. If this happens, we can safely conclude that LHS=RHS and hence proved
So, taking the RHS
⇒RHS=1−sin⁡2θsin⁡θ\Rightarrow RHS = \dfrac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}
Again, we will use another trigonometric property which states that
⇒1−sin⁡2θ=cos⁡2θ\Rightarrow 1 - {\sin ^2}\theta = {\cos ^2}\theta
Replacing the value of 1−sin⁡2θ1 - {\sin ^2}\theta from the given property, our new RHS will be
cos⁡2θsin⁡θ=cos⁡θ×cos⁡θsin⁡θ\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }} = \cos \theta \times \dfrac{{\cos \theta }}{{\sin \theta }}
But at the starting we learnt that cos⁡θsin⁡θ=cot⁡θ\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta
So, when we replace cos⁡θsin⁡θ\dfrac{{\cos \theta }}{{\sin \theta }}by cot⁡θ\cot \theta . We will get
cos⁡θ×cos⁡θsin⁡θ=cos⁡θcos⁡θ =LHS  \cos \theta \times \dfrac{{\cos \theta }}{{\sin \theta }} = \cos \theta \cos \theta \\\ = LHS \\\
Hence, when we solve the RHS of the equation, we end up with a value which was our LHS. This thing proves that RHS=LHS and hence the question given to us is finally proved.

Note: For solving this question we have to be familiar with the trigonometry functions and their different forms and their inverse forms and also, we should be familiar with all the trigonometric identities such that the equations of the questions get simplified.