Question

# Evaluate the following : $\left(1+\mathrm{tan}A+\mathrm{sec}A\right)\left(1+\mathrm{cot}A-\mathrm{cosec}A\right).$

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Solution

## Evaluate $\left(1+\mathrm{tan}A+\mathrm{sec}A\right)\left(1+\mathrm{cot}A-\mathrm{cosec}A\right).$Convert the given expression into $\mathrm{sine}\mathrm{and}\mathrm{cosine}$$\begin{array}{rcl}\left(1+\mathrm{tan}A+\mathrm{sec}A\right)\left(1+\mathrm{cot}A-\mathrm{cosec}A\right)& =& \left(1+\frac{\mathrm{sin}A}{\mathrm{cos}A}+\frac{1}{\mathrm{cos}A}\right)\left(1+\frac{\mathrm{cos}A}{\mathrm{sin}A}-\frac{1}{\mathrm{sin}A}\right)\left[\because \mathrm{tan}A=\frac{\mathrm{sin}A}{\mathrm{cos}A},\mathrm{sec}A=\frac{1}{\mathrm{cos}A},\mathrm{cot}A=\frac{\mathrm{cos}A}{\mathrm{sin}A},\mathrm{cosec}\mathrm{A}=\frac{1}{\mathrm{sin}A}\right]\\ & =& \left(\frac{\mathrm{cos}A+\mathrm{sin}A+1}{\mathrm{cos}A}\right)\left(\frac{\mathrm{sin}A+\mathrm{cos}A-1}{\mathrm{sin}A}\right)\\ & =& \frac{{\left(\mathrm{sin}A+\mathrm{cos}A\right)}^{2}-1}{\mathrm{sin}A·\mathrm{cos}A}\left[\because \left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}\right]\\ & =& \frac{{\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A+2\mathrm{sin}A·\mathrm{cos}A-1}{\mathrm{sin}A·\mathrm{cos}A}\left[\because {\left(a+b\right)}^{2}={a}^{2}+{b}^{2}+2ab\right]\\ & =& \frac{1+2\mathrm{sin}A·\mathrm{cos}A-1}{\mathrm{sin}A·\mathrm{cos}A}\\ & =& \frac{2\mathrm{sin}A·\mathrm{cos}A}{\mathrm{sin}A·\mathrm{cos}A}\\ & =& 2\end{array}$Hence, the required answer is $\left(1+\mathrm{tan}A+\mathrm{sec}A\right)\left(1+\mathrm{cot}A-\mathrm{cosec}A\right)=2$.

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