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ddd [48]
3 years ago
5

Solve this query plzzz​

Mathematics
1 answer:
Olin [163]3 years ago
7 0

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 \\

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