Question

Prove the following expression.
$1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=\operatorname{cosec}\alpha $

Answer 1

Hint: We solve this problem by first considering the LHS of the given expression. Then we consider the formulas $\operatorname{cosec}\alpha =\dfrac{1}{\sin \alpha }$ and $\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }$ to convert the expression into sine and cosines. Then we substitute these formulas in the given expression and simplify it. Then we consider the formula ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ and use it to simplify it further. Then we factorise the numerator using the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and cancel the terms that are common to both numerator and denominator. Then we calculate the remaining value and find the answer.

Complete step-by-step solution
We are asked to prove that $1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=\operatorname{cosec}\alpha $.
So, let us consider the left-hand side expression of the above equation.
$\Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }$
Now let us consider the trigonometric formulas,
$\operatorname{cosec}\alpha =\dfrac{1}{\sin \alpha }$
$\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }$
Using this formula, we can write the LHS that we have considered above as,
\[\begin{align}
  & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{{{\left( \dfrac{\cos \alpha }{\sin \alpha } \right)}^{2}}}{1+\dfrac{1}{\sin \alpha }} \\
 & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }}{\dfrac{\sin \alpha +1}{\sin \alpha }} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{\left( {{\cos }^{2}}\alpha \right)\sin \alpha }{{{\sin }^{2}}\alpha \left( \sin \alpha +1 \right)} \\
 & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{{{\cos }^{2}}\alpha }{\sin \alpha \left( \sin \alpha +1 \right)} \\
\end{align}\]
Now let us consider the trigonometric identity formula,
${{\sin }^{2}}A+{{\cos }^{2}}A=1$
Using this we can convert the formula as,
${{\cos }^{2}}A=1-{{\sin }^{2}}A$
Using this formula, we can write the above equation as,
\[\Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{1-{{\sin }^{2}}\alpha }{\sin \alpha \left( \sin \alpha +1 \right)}\]
Now let us consider the formula,
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Using this formula, we can write the numerator of the above equation as,
\[\Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{\left( 1-\sin \alpha \right)\left( 1+\sin \alpha \right)}{\sin \alpha \left( \sin \alpha +1 \right)}\]
As we see the above equation, we can cancel the term \[\left( 1+\sin \alpha \right)\] in the numerator with the term \[\left( \sin \alpha +1 \right)\] in the denominator as both are same.
So, by cancelling them we get,
\[\begin{align}
  & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{1-\sin \alpha }{\sin \alpha } \\
 & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=\dfrac{\sin \alpha +1-\sin \alpha }{\sin \alpha } \\
 & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=\dfrac{1}{\sin \alpha } \\
\end{align}\]
As we know that $\operatorname{cosec}\alpha =\dfrac{1}{\sin \alpha }$, we can write the above equation as,
\[\Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=\operatorname{cosec}\alpha \]
Hence Proved.

Note: We can also solve this question in an alternative simpler method.
First, let us consider the left-hand side expression of the given equation.
$\Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }$
Now let us consider the trigonometric formulas,
${{\operatorname{cosec}}^{2}}\alpha -{{\cot }^{2}}\alpha =1$
We can also write the formula as,
${{\cot }^{2}}\alpha ={{\operatorname{cosec}}^{2}}\alpha -1$
Then using this we can write the above equation as,
$\Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{{{\operatorname{cosec}}^{2}}\alpha -1}{1+\operatorname{cosec}\alpha }$
Now let us consider the formula,
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Using this formula, we can write the numerator of the above equation as,
\[\Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\dfrac{\left( \operatorname{cosec}\alpha -1 \right)\left( \operatorname{cosec}\alpha +1 \right)}{1+\operatorname{cosec}\alpha }\]
As we see the above equation, we can cancel the term \[\left( \operatorname{cosec}\alpha +1 \right)\] in the numerator with the term \[1+\operatorname{cosec}\alpha \] in the denominator as both are same.
So, by cancelling them we get,
\[\begin{align}
  & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=1+\operatorname{cosec}\alpha -1 \\
 & \Rightarrow 1+\dfrac{{{\cot }^{2}}\alpha }{1+\operatorname{cosec}\alpha }=\operatorname{cosec}\alpha \\
\end{align}\]
Hence Proved.