
How do you verify the identity \[\tan \left( \dfrac{A}{2} \right)=\dfrac{\sin A}{1+\cos A}\]?
Answer
555.3k+ views
Hint: We will verify the identity using trigonometric identities. We start solving from RHS by substituting the identities we have. We again simplify the expression until we arrive at the solution where we get LHS .
Complete step-by-step solution:
So we start solving from RHS
\[\dfrac{\sin A}{1+\cos A}\]
We have identities
\[\sin 2\theta =2\sin \theta \cos \theta \] and \[\cos 2\theta =2{{\cos }^{2}}\theta -1\].
Now we substitute \[\dfrac{A}{2}\] in place of \[\theta \].
Then the two identities will look like
\[\sin 2\left( \dfrac{A}{2} \right)=2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)\]
And
\[\cos 2\left( \dfrac{A}{2} \right)=2{{\cos }^{2}}\left( \dfrac{A}{2} \right)-1\]
By simplifying them we will get
\[\sin A=2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)\]
And
\[\cos A=2{{\cos }^{2}}\left( \dfrac{A}{2} \right)-1\]
Now we substitute these identities in the RHS part.
RHS= \[\dfrac{\sin A}{1+\cos A}\]
Now we substitute above sin A and cos A in our RHS.
By substituting we will get
\[\Rightarrow \dfrac{2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}{1+2{{\cos }^{2}}\left( \dfrac{A}{2} \right)-1}\]
Now we have to simplify the expression.
By simplifying we will get
\[\Rightarrow \dfrac{2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}{2{{\cos }^{2}}\left( \dfrac{A}{2} \right)}\]
Now we have \[\cos \left( \dfrac{A}{2} \right)\] on numerator and denominator. So we cancel them both in numerator and denominator we will get
\[\Rightarrow \dfrac{2\sin \left( \dfrac{A}{2} \right)}{2\cos \left( \dfrac{A}{2} \right)}\]
We can cancel 2 on both numerator and denominator. We will get
\[\Rightarrow \dfrac{\sin \left( \dfrac{A}{2} \right)}{\cos \left( \dfrac{A}{2} \right)}\]
We know that \[\dfrac{\sin A}{\cos A}=\tan A\]
So using this formula we will get our expression as
\[\Rightarrow \tan \left( \dfrac{A}{2} \right)\]
So we got LHS we can say the identity \[\tan \left( \dfrac{A}{2} \right)=\Rightarrow \dfrac{\sin \left( \dfrac{A}{2} \right)}{\cos \left( \dfrac{A}{2} \right)}\]is verified.
Note: We can also do it starting from LHS also. We know tan A formula we substitute the formula and then we apply the above derived formulas in place of Sin A and cos A. Then by simplifying the terms we will arrive at RHS as above. Then we can say that the identity the given is verified. So to solve this question we should be aware of trigonometric identities.
Complete step-by-step solution:
So we start solving from RHS
\[\dfrac{\sin A}{1+\cos A}\]
We have identities
\[\sin 2\theta =2\sin \theta \cos \theta \] and \[\cos 2\theta =2{{\cos }^{2}}\theta -1\].
Now we substitute \[\dfrac{A}{2}\] in place of \[\theta \].
Then the two identities will look like
\[\sin 2\left( \dfrac{A}{2} \right)=2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)\]
And
\[\cos 2\left( \dfrac{A}{2} \right)=2{{\cos }^{2}}\left( \dfrac{A}{2} \right)-1\]
By simplifying them we will get
\[\sin A=2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)\]
And
\[\cos A=2{{\cos }^{2}}\left( \dfrac{A}{2} \right)-1\]
Now we substitute these identities in the RHS part.
RHS= \[\dfrac{\sin A}{1+\cos A}\]
Now we substitute above sin A and cos A in our RHS.
By substituting we will get
\[\Rightarrow \dfrac{2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}{1+2{{\cos }^{2}}\left( \dfrac{A}{2} \right)-1}\]
Now we have to simplify the expression.
By simplifying we will get
\[\Rightarrow \dfrac{2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}{2{{\cos }^{2}}\left( \dfrac{A}{2} \right)}\]
Now we have \[\cos \left( \dfrac{A}{2} \right)\] on numerator and denominator. So we cancel them both in numerator and denominator we will get
\[\Rightarrow \dfrac{2\sin \left( \dfrac{A}{2} \right)}{2\cos \left( \dfrac{A}{2} \right)}\]
We can cancel 2 on both numerator and denominator. We will get
\[\Rightarrow \dfrac{\sin \left( \dfrac{A}{2} \right)}{\cos \left( \dfrac{A}{2} \right)}\]
We know that \[\dfrac{\sin A}{\cos A}=\tan A\]
So using this formula we will get our expression as
\[\Rightarrow \tan \left( \dfrac{A}{2} \right)\]
So we got LHS we can say the identity \[\tan \left( \dfrac{A}{2} \right)=\Rightarrow \dfrac{\sin \left( \dfrac{A}{2} \right)}{\cos \left( \dfrac{A}{2} \right)}\]is verified.
Note: We can also do it starting from LHS also. We know tan A formula we substitute the formula and then we apply the above derived formulas in place of Sin A and cos A. Then by simplifying the terms we will arrive at RHS as above. Then we can say that the identity the given is verified. So to solve this question we should be aware of trigonometric identities.
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