an interesting question

(√sec x-tanx/√secx + tanx)
this whole under root of (secx - tanx/sec x + tanx)

18 Answers

1
Pavithra Ramamoorthy ·

;)

62
Lokesh Verma ·

yes 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 :)

1
coolpal ·

D ANS IS modesecx X (secx-tanx)

1
MATRIX ·

yaaa ankit

1
ankit mahapatra ·

But he did the derivative of y= tanx -secx

1
MATRIX ·

Deriative:

y=secx-tanx

dy/dx=sec2x-secxtanx

dy/dx=secx(secx-tanx) Done Anirudh[4][9][9]

11
Anirudh Narayanan ·

Done [4]

1
ankit mahapatra ·

The derivative is :

secxtanx - sec2x
or
secx(tanx-secx)

1
coolpal ·

FORGOT 2 ADD
v r supposed 2 differentiate it

sorry guys

plz bear wid it

1
Pavithra Ramamoorthy ·

secx-tanx...................

11
Anirudh Narayanan ·

[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]

1
Pavithra Ramamoorthy ·

hey i think dats all we ve to do....dis frm 10th basics..............

11
Anirudh Narayanan ·

Machan, she has not given what to do first........so we don't know what to do next... [7]

PS
I am not joking [4]

1
Pavithra Ramamoorthy ·

don joke da.. tell wat to do next...................

11
Anirudh Narayanan ·

I think we must integrate it [7]

1357
Manish Shankar ·

(√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?

1
Pavithra Ramamoorthy ·

if i am wron pls forgive..

1
Pavithra Ramamoorthy ·

or (1-sinx)/cosx

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