Question

Prove that (1+tan square A) (cot A) /cosec square A=tan A

Answer 1

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Answer 2

$\frac{(1+\text{tan}^{2} A)(\text{cot A})}{\text{cosec}^{2} A } = \text{tan A}$

Step-by-step explanation:

We are prove the following trigonometric expression below;

$\frac{(1+\text{tan}^{2} A)(\text{cot A})}{\text{cosec}^{2} A } = \text{tan A}$

Taking the left-hand of the expression;

$\frac{(1+\text{tan}^{2} A)(\text{cot A})}{\text{cosec}^{2} A }$  

As we know that $1+\text{tan}^{2} A = \text{sec}^{2} A$, $\text{cot A} = \frac{1}{\text{tan A}}$  and  $\text{cosec A} = \frac{1}{\text{sin A}}$, so using these we get;

$\frac{\text{sec}^{2} A \times \frac{1}{\text{tan A}} }{\frac{1}{\text{sin}^{2} A } }$  =  $\frac{\frac{1}{\text{cos}^{2} A} \times \frac{1}{\text{tan A}} }{\frac{1}{\text{sin}^{2} A } }$      {as $\text{sec A} = \frac{1}{\text{cos A}}$ }

                  =  $\frac{1}{\text{cos}^{2} A} \times \frac{1}{\text{tan A}} } \times {\frac{\text{sin}^{2} A}{1 } }$

                  =  $\frac{\text{sin}^{2} A}{\text{cos}^{2} A} \times \frac{1}{\text{tan A}} }$

Now, as we know that $\text{tan A} = \frac{\text{sin A}}{\text{cos A}}$, so;

    =  $\text{tan}^{2} A \times \frac{1}{\text{tan A}} }$  

    =  tan A = RHS

Hence, it is proved that $\frac{(1+\text{tan}^{2} A)(\text{cot A})}{\text{cosec}^{2} A } = \text{tan A}$ .