# Is `1/(sin^2x) + 1/(cos^2x) = (sec^2x)(csc^2x)` an identity ?

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To prove, let's try to express left side in terms of secant and cosecant.

To do so, express left side with same denominator.

`1/(sin^2x)*(cos^2x)/(cos^2x) + 1/(cos^2x)*(sin^2x)/(sin^2x)= sec^2x csc^2x`

`(cos^2x)/(cos^2xsin^2x)+(sin^2x)/(cos^2xsin^2x)= sec^2x csc^2x`

Add the fractions.

`(cos^2x + sin^2x)/(cos^2xsin^2x)= sec^2x csc^2x`

Base on the Pythagorean identity, `cos^2x + sin^2x = 1` .

`1/(cos^2xsin^2x)= sec^2x csc^2x`

Note that the reciprocal of sine is cosecant `( cscx= 1/(sinx))` ` `. And, the reciprocal of cosine is secant `(secx=1/(cosx))` .

`sec^2x csc^2x=sec^2x csc^2x`

Expressing left side in terms of secant and cosecant leads to same the expression at the right right side of the equation.

**Hence, this proves that `1/(sin^x) + 1/(cos^2x)= sec^2xcsc^2x` **.

It has to be determined if `1/(sin^2x) + 1/(cos^2x) = sec^2x*csc^2x` is an identity.

Start with the left hand side:

`1/(sin^2x) + 1/(cos^2x)`

=> `(cos^2x)/(sin^2x* cos^2x) + (sin^2x)/(sin^2x*cos^2x)`

=> `(cos^2x + sin^2x)/(sin^2x*cos^2x)`

use the identity `cos^2x + sin^2x = 1`

=> `1/(sin^2x*cos^2x)`

=> `1/(sin^2x)*1/(cos^2x)`

As `1/sin x = csc x` and `1/cos x = sec x`

=> `sec^2x*csc^2x`

This is the right hand right.

**This proves that `1/(sin^2x) + 1/(cos^2x) = sec^2x*csc^2x ` is an identity and is true for all values of x.**

1/sin^2x + 1/cos^2x = sec^2x. csc^2x

L.H.S :-

1/(sin^2x) + 1/(cos^2x)= (sin^2x + cos^2x)/(sin^2x.cos^2x)

=>1/(sin^2x)+1/(cos^2x)= 1/(sin^2x.cos^2x) [ since sin^2x + cos^2x = 1 ]

=> 1/(sin^2x) + 1/(cos^2x)= (1/sin^2x).(1/cos^2x)

=> 1/(sin^2x) + 1/(cos^2x)= csc^2x. sec^2x [ since 1/sin^2x = csc^2x and 1/cos^2x= sec^x ]

=> 1/(sin^2x) + 1/(cos^2x) = sec^2x . csc^2x

hence,

1/(sin^2x) + 1/(cos^2x) = sec^2x . csc^2x <--proved