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.