Question

prove that cos2A=1-tan^2A/1+tan^2A​

Answer 1

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Answer 2

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

To find: We have to prove that  $cos2A=\frac{1-tan^2A}{1+tan^2A}$

Solution:

To prove LHS = RHS we need to apply the basic formulas of trigonometry,

The double angle formula for cosine, sin, and tangent is:

$sin2A=2sinAcosA=\frac{2tanA}{1+tan^2A}$

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

$tan2A=\frac{2tanA}{1-tan^2A}$

To prove the above identity we need to start from the term $\frac{1-tan^2A}{1+tan^2A}$ so that it can be simplified to get the LHS.

∴Taking the term in RHS,

$\frac{1-tan^2A}{1+tan^2A}\\ = \frac{1-\frac{sin^2A}{cos^2A} }{1+\frac{sin^2A}{cos^2A} } \\= \frac{\frac{cos^2A-sin^2A}{cos^2A} }{\frac{cos^2A+sin^2A}{cos^2A} }\\= \frac{cos^2A-sin^2A}{cos^2A+sin^2A}\\= \frac{cos2A}{1} (cos^2A-sin^2A=cos2A, sin^2A+cos^2A=1)\\=cos2A$

= LHS

Final answer:

Hence it is verified that $cos2A=\frac{1-tan^2A}{1+tan^2A}$