Question

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

Answer 1

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,

& \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{aligned}$$ $$\begin{aligned} & \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{aligned}$$ Now let us consider the trigonometric identity formula, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ Using this we can convert the formula as, ${{\cos }^{2}}A=1-{{\sin }^{2}}A$ Using this formula, we can write the above equation as, $$\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{aligned} & \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{aligned}$$ As we know that $\operatorname{cosec}\alpha =\dfrac{1}{\sin \alpha }$, we can write the above equation as, $$\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{aligned} & \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{aligned}$$ Hence Proved.