1The derivative is :
secxtanx - sec2x
or
secx(tanx-secx)
(√sec x-tanx/√secx + tanx)
this whole under root of (secx - tanx/sec x + tanx)
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18 Answers
62yes coolpal if u see closely,
the answer is kind of myopic
the answer is correct..
the mistake is that secx-tanx is always positive otherwise the above expression that you gave will be not defined.
So what that will do is that it will make the answer given by manish and anirudh same as the answer given in the book used by you :)
1Deriative:
y=secx-tanx
dy/dx=sec2x-secxtanx
dy/dx=secx(secx-tanx) Done Anirudh[4][9][9]
11[9]
I think we have to integrate da
ok we'll do everything we can with this [4]
secx - tanx
=(1-sinx)/cosx
= 1-tan(x/2)/1+tan(x/2)
derivative is
secx(secx-tanx)
Integral is
I = ∫(secx - tanx) dx
= ∫(secx - secxtanx/secx) dx
= log(secx+tanx) - logsecx
= log(1+sinx) [4]
1hey i think dats all we ve to do....dis frm 10th basics..............
11Machan, she has not given what to do first........so we don't know what to do next... [7]
PS
I am not joking [4]
1357(√sec x-tanx/√secx + tanx)
multiply by √sec x-tanx
so we get
(sec x-tanx)/(√sec2x-tan2x)
= (sec x-tanx)
now what do we need to find?