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 \[\cos \theta \].