Question

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

Answer 1

Hint: The very first step to solve this problem, replace ${\cos ^2}x$with $\dfrac{1}{{\sin \theta }}$. After this take the $\sin \theta $on the LHS and take it to RHS. After this, solve and simplify the RHS. You will get a term $1 - {\sin ^2}x$on the RHS. Replace this term by ${\cos ^2}x$. 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
\[
  \cot \theta = \dfrac{1}{{\tan \theta }} \\
  \sin \theta = \dfrac{1}{{\csc \theta }} \\
  \cos \theta = \dfrac{1}{{\sec \theta }} \\
\]

Complete step-by-step solution:
The given question we have is\[\sin \theta + \cot \theta \cos \theta = \csc \theta \]
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:-
\[
  \cot \theta = \dfrac{1}{{\tan \theta }} \\
  \sin \theta = \dfrac{1}{{\csc \theta }} \\
  \cos \theta = \dfrac{1}{{\sec \theta }} \\
\]
Now, beginning to solve the problem, we will replace \[\csc \theta \]on the RHS to $\dfrac{1}{{\sin \theta }}$
Doing this we will end up with
$ \Rightarrow \sin \theta + \cot \theta \cos \theta = \dfrac{1}{{sin\theta }}$
Taking the $\sin \theta $to the RHS,
$
   \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
$ \Rightarrow RHS = \dfrac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}$
Again, we will use another trigonometric property which states that
$ \Rightarrow 1 - {\sin ^2}\theta = {\cos ^2}\theta $
Replacing the value of $1 - {\sin ^2}\theta $from the given property, our new RHS will be
$\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }} = \cos \theta \times \dfrac{{\cos \theta }}{{\sin \theta }}$
But at the starting we learnt that $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $
So, when we replace $\dfrac{{\cos \theta }}{{\sin \theta }}$by $\cot \theta $. We will get
$
  \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.