Question

express tanA in terms of sin2A and cos2A​

Answer 1

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

We know that,

$\purple{ \boxed{ \bf{sin2x = 2sinxcosx}}}$

and

$\purple{ \boxed{ \bf{cos2x =1 - 2 {sin}^{2}x }}}$

Now, Consider

$\rm :\longmapsto\:tanA$

$\rm \: = \: \dfrac{sinA}{cosA}$

On multiply and divide by 2 sinA, we get

$\rm \: = \: \dfrac{ {2sin}^{2} A}{2sinA \: cosA}$

$\rm \: = \: \dfrac{1 - cos2A}{sin2A}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \purple{ \underbrace{ \boxed{ \bf\ \: \: \: \: \: tanA = \dfrac{1 - cos2A}{sin2A} \: \: \: \: \: \: \: }}}$

Additional Information :-

Let's solve the same type of problem!!!!

Question :- Express cotA in terms of sin2A and cos2A.

Solution :-

$\rm :\longmapsto\:cotA$

$\rm \: = \: \dfrac{cosA}{sinA}$

On multiply and divide by cosA, we get

$\rm \: = \: \dfrac{ {2cos}^{2} A}{2sinA \: cosA}$

$\rm \: = \: \dfrac{1 + cos2A}{sin2A}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \purple{ \underbrace{ \boxed{ \bf\ \: \: \: \: \: cotA = \dfrac{1 + cos2A}{sin2A} \: \: \: \: \: \: \: }}}$

Know more about Formula's

$\purple{ \boxed{ \bf{sin2x = 2sinxcosx = \frac{2tanx}{1 + {tan}^{2}x}}}}$

$\purple{ \boxed{ \bf{cos2x = {cos}^{2}x - {sin}^{2}x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1}}}$

$\purple{ \boxed{ \bf{cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x}}}}$

$\purple{ \boxed{ \bf{tan2x = \frac{2tanx}{1 - {tan}^{2} x}}}}$

$\purple{ \boxed{ \bf{1 - cos2x = {2sin}^{2}x}}}$

$\purple{ \boxed{ \bf{1 + cos2x = {2cos}^{2}x}}}$