Question

Prove the following statement:
\[\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A-1}+\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A+1}=2{{\sec }^{2}}A\]

Answer 1

Hint: Consider the LHS of the given equation and take (cosec A – 1) (cosec A + 1) as LCM and simplify the given expression. Now use \[{{\operatorname{cosec}}^{2}}\theta -1={{\cot }^{2}}\theta \] and \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] and \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\] to prove the desired result.
Complete step-by-step answer:
Here, we have to prove that \[\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A-1}+\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A+1}=2{{\sec }^{2}}A\]
Let us consider the LHS of the expression given in the question.
\[E=\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A-1}+\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A+1}\]
By taking the LCM as (cosec A – 1) (cosec A + 1) and simplifying the above expression, we get,
\[E=\dfrac{\left( \operatorname{cosec}A \right)\left( \operatorname{cosec}A+1 \right)+\left( \operatorname{cosec}A \right)\left( \operatorname{cosec}A-1 \right)}{\left( \operatorname{cosec}A-1 \right)\left( \operatorname{cosec}A+1 \right)}\]
We know that \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]. By using this in the denominator of the above expression, we get,
\[E=\dfrac{\left( \operatorname{cosec}A \right)\left( \operatorname{cosec}A+1 \right)+\operatorname{cosec}A\left( \operatorname{cosec}A-1 \right)}{\left( {{\operatorname{cosec}}^{2}}A-1 \right)}\]
By simplifying the above expression, we get,
\[E=\dfrac{{{\operatorname{cosec}}^{2}}A+\operatorname{cosec}A+{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}{\left( {{\operatorname{cosec}}^{2}}A-1 \right)}\]
By canceling cosec A from the numerator of the above expression, we get,
\[E=\dfrac{{{\operatorname{cosec}}^{2}}A+{{\operatorname{cosec}}^{2}}A}{\left( {{\operatorname{cosec}}^{2}}A-1 \right)}\]
\[E=\dfrac{2{{\operatorname{cosec}}^{2}}A}{\left( {{\operatorname{cosec}}^{2}}A-1 \right)}\]
We know that \[{{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =1\] or \[{{\operatorname{cosec}}^{2}}\theta -1={{\cot }^{2}}\theta \]. By using this in the denominator of the above expression, we get,
\[E=\dfrac{2{{\operatorname{cosec}}^{2}}A}{{{\cot }^{2}}A}\]
Now, we know that \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] and \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]
By using this in the above expression, we get,
\[E=\dfrac{2{{\left( \dfrac{1}{\sin A} \right)}^{2}}}{{{\left( \dfrac{\cos A}{\sin A} \right)}^{2}}}\]
\[E=\dfrac{\dfrac{2}{{{\sin }^{2}}A}}{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}}\]
\[E=\left( \dfrac{2}{{{\sin }^{2}}A} \right)\times \left( \dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A} \right)\]
By canceling the like terms from the above equation, we get,
\[E=\dfrac{2}{{{\cos }^{2}}A}\]
We know that \[\cos \theta =\dfrac{1}{\sec \theta }\]. By using this in the above expression, we get,
\[E=\dfrac{2}{{{\left( \dfrac{1}{\sec A} \right)}^{2}}}\]
\[E=2{{\left( \sec A \right)}^{2}}\]
\[E=2{{\sec }^{2}}A\]
E = RHS
So, we get, LHS = RHS
Hence proved
So, we have proved that \[\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A-1}+\dfrac{\operatorname{cosec}A}{\operatorname{cosec}A+1}=2{{\sec }^{2}}A\]

Note: Some students are often uncomfortable with \[\operatorname{cosec}\theta \] or \[\sec \theta \] or \[\cot \theta \] that are the reciprocals. So, in that case, students could also convert the given LHS terms of \[\sin \theta \] by writing \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] and then solving the expression in the same way in terms of \[\sin \theta \] and \[\cos \theta \] to get the desired result. Students should always learn the formula in terms of \[\operatorname{cosec}\theta ,\cot \theta \] and \[\sec \theta \] also and not only in terms of \[\sin \theta ,\cos \theta ,\tan \theta \] to solve the question faster and easily.