Question

Prove that Cos2A=1-tanA\1+tanA

Answer 1

Appropriate Question :-

Prove that .

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

$\large\underline{\bf{Solution-}}$

Consider,

$\rm :\longmapsto\:cos2A$

can be rewritten as

$\rm \: = \: cos(A + A)$

We know,

$\boxed{ \sf{cos(x + y) = cosxcosy - sinxsiny}}$

Using this identity, we get

$\rm \: = \: cosAcosA - sinAsinA$

$\rm \: = \: {cos}^{2}A - {sin}^{2}A$

can be rewritten as

$\rm \: = \: \dfrac{ {cos}^{2} A - {sin}^{2}A }{1}$

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

$\rm \: = \: \dfrac{ {cos}^{2}A\bigg(1 - \dfrac{ {sin}^{2} A}{ {cos}^{2} A} \bigg) }{{cos}^{2}A\bigg(1 + \dfrac{ {sin}^{2} A}{ {cos}^{2} A} \bigg)}$

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

Hence,

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

Additional Information

Let's solve the same type of problem!!

Question :- Prove that,

$\boxed{ \sf{cos2A = 1 - 2 {sin}^{2}A = {2cos}^{2}A - 1}}$

Solution :-

Consider,

$\rm :\longmapsto\:cos2A$

can be rewritten as

$\rm \: = \: cos(A + A)$

$\rm \: = \: cosAcosA - sinAsinA$

$\rm \: = \: {cos}^{2}A - {sin}^{2}A$

$\rm \: = \: 1 - {sin}^{2}A - {sin}^{2}A$

$\bf \: = \: 1 -2 {sin}^{2}A$

$\rm \: = \: 1 -2(1 - {cos}^{2}A)$

$\rm \: = \: 1 -2 + 2{cos}^{2}A$

$\bf\: = \: 2{cos}^{2}A - 1$

Hence,

$\boxed{ \bf{cos2A \: = \: 1 - 2 {sin}^{2}A \: = \: {2cos}^{2}A - 1}}$