Question

prove that 2cos^2A=1-tan^2A÷1+tan^2A​

Answer 1

Answer:

$RHS =\frac{1-tan^{2}A}{1+tan^{2}A}\\=\frac{1-\frac{sin^{2}A}{cos^{2}A}}{1+\frac{sin^{2}A}{cos^{2}A}}$

$=\frac{\frac{cos^{2}A-sin^{2}A}{cos^{2}A}}{\frac{cos^{2}A+sin^{2}A}{cos^{2}A}}$

$=\frac{cos^{2}A-sin^{2}A}{cos^{2}A+sin^{2}A}\\=\frac{cos2A}{1}\\=cos2A\\=LHS$

Therefore,

$\frac{1-tan^{2}A}{1+tan^{2}A}=cos2A$

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Answer 2

Answer:

Step-by-step explanation: