Question

Prove the following statement:
$\dfrac{1}{\cot A+\tan A}=\sin A\cos A$

Answer 1

Hint: First of all, consider the LHS of the given equation and use $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and convert the whole expression in terms of sin A and cos A. Now, simplify the given expression and substitute ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ to prove the desired result.
Complete step-by-step answer:
In this question, we have to prove that $\dfrac{1}{\cot A+\tan A}=\sin A\cos A$. Let us consider the LHS of the equation given in the question.
$E=\dfrac{1}{\cot A+\tan A}$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. By using these in the denominator of the above expression, we get,
$E=\dfrac{1}{\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A}}$
By taking $\sin A\cos A$ as LCM in the denominator and simplifying the expression, we get,
$E=\dfrac{1}{\dfrac{\cos A.\cos A+\sin A.\sin A}{\sin A\cos A}}$
$E=\dfrac{1}{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{\sin A\cos A}}$
$E=\dfrac{\cos A\sin A}{{{\cos }^{2}}A+{{\sin }^{2}}A}$
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$. By using this in the above expression, we get,
$E=\dfrac{\cos A\sin A}{1}$
$E=\cos A\sin A$
E = RHS
So, we get, LHS = RHS
Hence proved.
So, we have proved that $\dfrac{1}{\cot A+\tan A}=\sin A\cos A$.

Note: In this question, students often make mistakes while taking the LCM. After taking LCM, they often write $\dfrac{\cos A+\sin A}{\cos A\sin A}$ which is wrong. The right expression would be $\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{\cos A\sin A}$. So, this must be taken care of. Also, students must remember the general trigonometric formulas like ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$. Also, it is always reliable to convert the whole expression in terms of $\sin \theta$ and $\cos \theta$.