Question

Solve this query plzzz​

Answer 1

Step-by-step explanation:

$\frac{\sin \:4A}{\cos \:2A} \times \frac{1 - \cos \:2A}{1 - \cos\:4A} = \tan \:A \\ \\ LHS = \frac{\sin \:4A}{\cos \:2A} \times \frac{1 - \cos \:2A}{1 - \cos\:4A} \\ \\ = \frac{2\sin \:2A.\cos \:2A}{\cos \:2A} \times \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\ \\ = 2\sin \:2A \times \frac{1 - 2 { \cos}^{2}A + 1}{1 - 2 { \cos}^{2}2A + 1 } \\ \\ = 2\sin \:2A \times \frac{2- 2 { \cos}^{2}A }{2 - 2 { \cos}^{2}2A } \\ \\ = 2\sin \:2A \times \frac{2(1 - { \cos}^{2}A) }{2 (1- { \cos}^{2}2A) } \\ \\ = 2\sin \:2A \times \frac{1 - { \cos}^{2}A}{1- { \cos}^{2}2A } \\ \\ = 2\sin \:2A \times \frac{ { \sin}^{2}A}{{ \sin}^{2}2A } \\ \\ = 2 \times \frac{ { \sin}^{2}A}{{ \sin}2A } \\ \\ = 2 \times \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \: A } \\ \\ = \frac{ { \sin}A}{ \cos \: A } \\ \\ = tan \: A \\ \\ = RHS$