Question
Prove the following identity :
tanA+cotA=secAcosecA
tanA+cotA=secAcosecA
✓
Answer
$
\begin{aligned}
& \tan \mathrm{A}+\cot \mathrm{A}=\sec \mathrm{A} \operatorname{cosec} \mathrm{A} \\
& \text { Consider LHS }=\tan \mathrm{A}+\cot \mathrm{A} \\
& \tan \mathrm{A}+\cot \mathrm{A}=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin ^2 A+\cos ^2 A}{\sin A \cdot \cos A} \\
& \Rightarrow \tan A+\cot A=\frac{1}{\sin A \cdot \cos A}=\frac{1}{\sin A} \frac{1}{\cos A} \\
& \Rightarrow \tan \mathrm{A}+\cot \mathrm{A}=\operatorname{cosec} \mathrm{A} \cdot \sec \mathrm{A}=\mathrm{RHS}
\end{aligned}
$
\begin{aligned}
& \tan \mathrm{A}+\cot \mathrm{A}=\sec \mathrm{A} \operatorname{cosec} \mathrm{A} \\
& \text { Consider LHS }=\tan \mathrm{A}+\cot \mathrm{A} \\
& \tan \mathrm{A}+\cot \mathrm{A}=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin ^2 A+\cos ^2 A}{\sin A \cdot \cos A} \\
& \Rightarrow \tan A+\cot A=\frac{1}{\sin A \cdot \cos A}=\frac{1}{\sin A} \frac{1}{\cos A} \\
& \Rightarrow \tan \mathrm{A}+\cot \mathrm{A}=\operatorname{cosec} \mathrm{A} \cdot \sec \mathrm{A}=\mathrm{RHS}
\end{aligned}
$
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