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Definite Integral as Limit of Sum

Prove that ...

Question

Prove that $(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {tan}^{2} A$

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Solution

$L . H . S = \frac{1 + t a n^{2} A}{1 + c o t^{2} A}$

$= \frac{1 + t a n^{2} A}{1 + \frac{1}{t a n^{2} A}}$

$= \frac{(1 + t a n^{2} A) t a n^{2} A}{(t a n^{2} A + 1)}$

$∴ \frac{1 + t a n^{2} A}{1 + c o t^{2} A} = t a n^{2} A$

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\frac{tan (A + B)}{cot (A - B)} = \frac{{tan}^{2} A - {tan}^{2} B}{1 - {tan}^{2} A {tan}^{2} B}

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

\frac{{csc}^{2} A + {sec}^{2} A}{{csc}^{2} A - {sec}^{2} A} = \frac{1 + {tan}^{2} A}{1 - {tan}^{2} A}

tan A = \sqrt{2} - 1

\frac{tan A}{1 + {tan}^{2} A} = \frac{\sqrt{2}}{4}

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