plzzzz kal mera exam hai...
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astha1917:
best of luck for your exams:)
he is right soumika!!
ok (thanks by the way astha and 111)
Answers
Answered by
3
LHS = (sinA-cosA+1)/(sinA+cos-1)
= (1 + sinA) (1 - sinA)/cos A(1 - sinA)
= 1 - sin2A/cos A(1 - sinA)
= cos2A/cos A(1 - sinA)
= cosA/(1 - sinA)
= 1/ (1/cosA - sinA/cosA)
= 1/(secA - tanA) = RHS
Hence Proved
= (1 + sinA) (1 - sinA)/cos A(1 - sinA)
= 1 - sin2A/cos A(1 - sinA)
= cos2A/cos A(1 - sinA)
= cosA/(1 - sinA)
= 1/ (1/cosA - sinA/cosA)
= 1/(secA - tanA) = RHS
Hence Proved
Answered by
5
heya friends!!!
from lhs
=>sin¢-cos¢-1
---------------------.
sin¢+cos¢-1. ● 【dividing by cos¢ on numerator and denomenator 】
then we get.....
tan¢-1+sec¢
=>--------------------
tan¢+1-sec¢
=>tan¢+sec¢-1
--------------------------------------------
tan¢+(sec^2¢-tan^2¢ ) -sec¢. ◆since.. 【sec^2¢-tan^2¢=1 】
*(tan¢+sec¢-1)
=>-----------------------------------
(sec¢-tan¢¢)(tan¢ +sec¢-1)
◆【here sec¢+tan¢-1 is common in denomenator 】●since it is cancelled here
remaining
=>1/sec¢-tan¢
hence Rhs .
1/sec¢-tan¢..Rhs .....prooved ...
hope it help you.
@rajukumar☺☺☺
from lhs
=>sin¢-cos¢-1
---------------------.
sin¢+cos¢-1. ● 【dividing by cos¢ on numerator and denomenator 】
then we get.....
tan¢-1+sec¢
=>--------------------
tan¢+1-sec¢
=>tan¢+sec¢-1
--------------------------------------------
tan¢+(sec^2¢-tan^2¢ ) -sec¢. ◆since.. 【sec^2¢-tan^2¢=1 】
*(tan¢+sec¢-1)
=>-----------------------------------
(sec¢-tan¢¢)(tan¢ +sec¢-1)
◆【here sec¢+tan¢-1 is common in denomenator 】●since it is cancelled here
remaining
=>1/sec¢-tan¢
hence Rhs .
1/sec¢-tan¢..Rhs .....prooved ...
hope it help you.
@rajukumar☺☺☺
sone dete hai unhe..
byee
thnkss
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