How do you prove $\dfrac{1-\cos x}{\sin x}=\tan \dfrac{x}{2}$ ?
Answer
588.6k+ views
Hint: We will use the trigonometric identities to solve the given expression. The correct way to solve this question is to find the relation between LHS and RHS by simplifying the LHS of the given expression. To prove $\dfrac{1-\cos x}{\sin x}=\tan \dfrac{x}{2}$ first we will consider the LHS of the given expression and then by using the half angle formula simplify the LHS and prove it equal to the RHS.
The following formulas we will use to simplify the equation:
$1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Complete step by step answer:
We have been given an expression $\dfrac{1-\cos x}{\sin x}=\tan \dfrac{x}{2}$.
We have to prove that LHS=RHS.
Let us consider the LHS of the given expression we have $\dfrac{1-\cos x}{\sin x}$
Now, as we have given a half angle in the RHS so we need to convert the LHS also into the half angle form.
Now, we know that $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$.
Now, substituting the above formula in the obtained equation we get
$\Rightarrow \dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{\sin x}$
Now, we know that $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$.
Now, substituting the above formula in the obtained equation we get
$\Rightarrow \dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$
Now, simplify the above obtained equation we get
$\Rightarrow \dfrac{2\sin \dfrac{x}{2}\sin \dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$
Now, cancel out the common terms we get
$\Rightarrow \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}$
Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
Now, substituting the above identity in the obtained equation we get
$\Rightarrow \tan \dfrac{x}{2}$
Hence we get $\dfrac{1-\cos x}{\sin x}=\tan \dfrac{x}{2}$
LHS=RHS
Hence proved
Note: To solve such types of questions students must have the knowledge of basic trigonometric ratios, trigonometric identities and trigonometric functions. Also students must know the conversion formula of one function into another to solve this type of question.
The following formulas we will use to simplify the equation:
$1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Complete step by step answer:
We have been given an expression $\dfrac{1-\cos x}{\sin x}=\tan \dfrac{x}{2}$.
We have to prove that LHS=RHS.
Let us consider the LHS of the given expression we have $\dfrac{1-\cos x}{\sin x}$
Now, as we have given a half angle in the RHS so we need to convert the LHS also into the half angle form.
Now, we know that $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$.
Now, substituting the above formula in the obtained equation we get
$\Rightarrow \dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{\sin x}$
Now, we know that $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$.
Now, substituting the above formula in the obtained equation we get
$\Rightarrow \dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$
Now, simplify the above obtained equation we get
$\Rightarrow \dfrac{2\sin \dfrac{x}{2}\sin \dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$
Now, cancel out the common terms we get
$\Rightarrow \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}$
Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
Now, substituting the above identity in the obtained equation we get
$\Rightarrow \tan \dfrac{x}{2}$
Hence we get $\dfrac{1-\cos x}{\sin x}=\tan \dfrac{x}{2}$
LHS=RHS
Hence proved
Note: To solve such types of questions students must have the knowledge of basic trigonometric ratios, trigonometric identities and trigonometric functions. Also students must know the conversion formula of one function into another to solve this type of question.
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