Prove that - 1 - tan A1 - A -2 - tan2 A-

To prove: [ 1 + tan #160;A1 + #160;A ]^2 = tan^2 A .LHS= [ 1 + tan #160;A1 + 1tan #160;A ]^2 .= [ (1 + tan #160;A)(1 + tan #160;A)tan ...

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Trigonometric Ratios for Sum of Two Angles

Prove that ...

Question

Prove that ${[\frac{1 + tan A}{1 + cot A}]}^{2} = {tan}^{2} A$ .

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tan A = \sqrt{2} - 1

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

(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{c o t A})^{2} = t a n^{2} A

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

sin (90^{\circ} - A) cos (90^{\circ} - A) = \frac{tan A}{1 + {tan}^{2} A} .

sin (90^{0} - A) . cos (90^{0} - A) = \frac{tan A}{1 + {tan}^{2} A}

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