Question

Prove the following statement:
\[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\]

Answer 1

Hint: First of all, consider the LHS of the given equation and convert the whole expression in terms of sin A and cos A by using the formula \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and simplify the expression. Now, divide numerator and denominator by sin A and use \[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \] to prove the desired result.
Complete step-by-step answer:
In this question, we have to prove that \[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\]. Let us consider the LHS of the equation given in the question.
\[E=\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. So by substituting the value of tan A in terms of sin A and cos A in the above expression, we get,
\[E=\dfrac{1+{{\left( \dfrac{\sin A}{\cos A} \right)}^{2}}}{1+{{\cot }^{2}}A}\]
We also know that \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]. So by substituting the value of cot A in terms of cos A and sin A in the above expression, we get,
\[E=\dfrac{1+{{\left( \dfrac{\sin A}{\cos A} \right)}^{2}}}{1+{{\left( \dfrac{\cos A}{\sin A} \right)}^{2}}}\]
\[E=\dfrac{1+\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}}{1+\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}}\]
By simplifying the above expression, we get,
\[E=\dfrac{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{{{\cos }^{2}}A}}{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{{{\sin }^{2}}A}}\]
By further simplification, we get,
\[E=\left( \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{{{\cos }^{2}}A} \right).\left( \dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A+{{\sin }^{2}}A} \right)\]
We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. By using this in the above expression, we get,
\[E=\left( \dfrac{1}{{{\cos }^{2}}A} \right).\left( \dfrac{{{\sin }^{2}}A}{1} \right)\]
\[E=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\]
E = RHS
So, we get, LHS = RHS
Hence proved.
So, we have proved that \[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\].

Note: Students can also solve this question in the following way. Let us consider the LHS of the given equation.
\[E=\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}\]
By substituting \[\tan A=\dfrac{1}{\cot A}\], we get,
\[E=\dfrac{1+\dfrac{1}{{{\cot }^{2}}A}}{1+{{\cot }^{2}}A}\]
\[E=\dfrac{\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}}{\left( 1+{{\cot }^{2}}A \right)}\]
\[E=\dfrac{\left( 1+{{\cot }^{2}}A \right)}{\left( {{\cot }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)}\]
Now by canceling the like terms and substituting \[\cot A=\dfrac{\cos A}{\sin A}\], we get,
\[E=\dfrac{1}{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}}\]
So, we get, \[E=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}=RHS\]
So, LHS = RHS
Hence proved.