Question

PROVE THAT 1-TAN2A/1+TAN2A=COS2A-SIN2A

Answer 1

SOLUTION

TO PROVE

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

PROOF

PROCESS : 1

$\displaystyle \sf{ \frac{1 - { \tan}^{2}A }{1 + { \tan}^{2}A} }$

$\displaystyle \sf{ = \frac{1 - \frac{{ \sin}^{2}A}{{ \cos}^{2}A} }{1 + \frac{{ \sin}^{2}A}{{ \cos}^{2}A}} }$

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

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

$\displaystyle \sf{ = \frac{{ \cos}^{2}A - { \sin}^{2}A}{1} }$

$\displaystyle \sf{ = { \cos}^{2}A - { \sin}^{2}A }$

PROCESS : 2

$\displaystyle \sf{ \frac{1 - { \tan}^{2}A }{1 + { \tan}^{2}A} }$

$\displaystyle \sf{ = \cos \:2A }$

$\displaystyle \sf{ = { \cos}^{2}A - { \sin}^{2}A}$

Hence proved

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Answer 2

Answer:

The proof

Step-by-step explanation:

LHS= 1-tan²A/1+tan²A

1-sin²A/cos²A

-------------------

sec²A

 1                    sin²A/cos²A

-------     -         -----------------

sec²A                 sec²A

                      sin²A/cos²A

cos²A    -        -----------------

                          1/cos²A

cos²A gets cancelled

= cos²A - sin²A = RHS

LHS = RHS Hence proved