Solve this question of maths
Question no.9 part 2nd
Attachments:
Answers
Answered by
2
Hey !!!
Q. no 9 ,part 2nd
Solution :-
From LHS
tan²x /secx - 1
sin²x /cos²x
--------------------
(1/cosx - 1 )²
sin²x /cos²x
---------------------
(1 - cosx )²/cos²x
sin²x /(1 - cosx )²
( 1 - cos²x ) / (1 - cosx)²
(1 - cosx ) ( 1 + cosx ) / ( 1 - cosx ) ( 1 + cosx )
=> (1 + cosx )/( 1 - cosx ) Rhs prooved ♻
________________________
Hope it helps you !!!
@Rajukumar111 (^^)
Q. no 9 ,part 2nd
Solution :-
From LHS
tan²x /secx - 1
sin²x /cos²x
--------------------
(1/cosx - 1 )²
sin²x /cos²x
---------------------
(1 - cosx )²/cos²x
sin²x /(1 - cosx )²
( 1 - cos²x ) / (1 - cosx)²
(1 - cosx ) ( 1 + cosx ) / ( 1 - cosx ) ( 1 + cosx )
=> (1 + cosx )/( 1 - cosx ) Rhs prooved ♻
________________________
Hope it helps you !!!
@Rajukumar111 (^^)
qqqqqqppppppqpqpqpqp:
yup it helps me a lot
Answered by
0
hope it helps you dear....
Attachments:
Similar questions