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Is this correct?

Let I=

(1/x) dx

using uv method :

u=1/x and dv=dx

so du= (-1/x²) dx and v=x

I=(1/x ) * ( x) -

( -1/x²) * (x) dx

I=1-

(-1/x) dx

I=1+

(1/x) dx

But I=

(1/x)dx

I= 1 + I

Therefore 0=1

Taking cos on both sides
cos y = cos ( cos^-1 (( 3+5cos x ) / ( 5+3cos x )))
cosy = ( 3+5cos x ) / ( 5+3cos x )
differentiating on both sides
( -sin y )( dy/dx ) = (( 5+3cos x )( 0+5(-sin x )) - ( 3+5cos x )( 0+3(-sin x )))/( 5+3cos x )^2.
( -sin y )( dy/dx ) = ( -16sin x ) / ( 5+3cos x )^2.
Therefore , ( dy/dx ) = ( 16sinx ) / ( (sin y) ( 5+3cos x )^2 )

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If f(x)=[x}

g(x)=sin

x , 0

x

2 . Find (g.f)(x)?

Find Domain?

f(x)=log_0.5(log((3x+2)/(3x-1))).

Find Domain?

f(x)=^{[2 ]}

((-log_x-3(x-1))(^{[2 ]}

-x²+2x+8)).

Find Domain?

f(x)=^{[2 ]}

((cosx-1/2)(6+35x-6x²))

If atan A+btanB=(a+b)tan((A+B)/2) then

a)

is isoceles.

b)

is equilateral.

c)

is right.

d)none of these.

In a

(b²-c²)cot A+(c²-a²)cot B+(a²-b²)cot C is

a) a+b+c.

b) abc.

c) 0.

d) none of these.

Prove that :

tan²(

/16) + tan²(2

/16)+tan²(3

/16)+tan²(4

/16)+tan²(5

/16)+tan²(6

/16)+tan²(7

/16) = 35 .