Prove that: cos A = (1 - tan² A/2) / (1 + tan² A/2) | Trigonometry - Sub Multiple Angles | SciPiPupil

Question: Prove that: $cos A = \dfrac{1 - tan^2 \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$

Solution:

To prove: $cos A = \dfrac{1 - tan^2 \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$

Taking LHS,

$= cos A$

$= \dfrac{cos A}{1}$

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

[Dividing numerator and denominator by cos²A]

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

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

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

= RHS proved.

Some SUB Multiple Angles Formulae

$sin A = 2 sin\frac{A}{2}cos\frac{A}{2}$

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

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

$cot A = \dfrac{cot^2 \frac{A}{2} -1}{2 cot \frac{A}{2}}$

$sin A = 3 sin \frac{A}{3} - 4sin^3 \frac{A}{3}$

$cos A = 4cos^3 \frac{A}{3} - 3 cos \frac{A}{3}$

$tan A = \dfrac{3tan \frac{A}{3} - tan^3 \frac{A}{3}}{ 1 - 3tan^2 \frac{A}{3}}$

$cot A = \dfrac{3cot \frac{A}{3} - cot^3 \frac{A}{3}}{ 1 - 3cot^2 \frac{A}{3}}$

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