We need to prove that:

sec^2 x +csec^2 x = (sec^2 x)(csec^2 x)

Let us start with the left side:

We know that secx= 1/cosx and csec x = 1/sinx

==> sec^2 x+ cosec^2 x= 1/cos^2 x + 1/sin^2 x

= (sen^2 x + cos^2 x)/(sin^2 x)(cos^2 x)

Now we know that sin^2 x + cos^2 x= 1

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

= (sec^2 x)*(csec^2 x)

sec^2 X + cosec^2 X = (sec^2 X)*(cosec^2 X)

L.H.S=sec^2 X + cosec^2 X

=(1/cos^2 X)+(1/sin^2 X)

=(sin^2 X+cos^2 X)/(sin^2 X*cos^2 X)

=1/(sin^2 X*cos^2 X) { because sin^2 X+cos^2 X=1 }

=(sec^2 X)*(cosec^2 X)

=R.H.S

T prove that sec^2x+cosec^2x = sec^2*cosec^2x.

Solution:

We start with the most popular trigonometric identity:

(sinx)^2+(cosx)^2 = 1. Divide both sides bt (cosx)^2(sinx)^2.

{(sinx)^2+(cosx)^2}/[(cosx)^2sinx)^2] = 1/[(cosx)^2(sinx)^2}^2. Or

(sinx)^2/{cosx)^2sinx)^2} + (cosx)^2/ {cosx)^2(sinx)^2} = (secx)^2(cosecx)^2. Or

(secx)^2 + (cosecx)^2 = (secx)^2 * (cosecx)^2

Prove that sec^2 A + cosec^2 A = (sec^2 A)*(cosec^2 A)

L.H.S = sec^2 A + cosec^2 A

= (1/cos^2 A)+(1/sin^2 A)

= (sin^2 A+cos^2 A)/(sin^2 X*cos^2 A)

= 1/(sin^2 A*cos^2 A) [ sin^2 A+cos^2 A=1 ]

= (sec^2 A)*(cosec^2 A) = R.H.S

L.H.S = R.H.S

Hence Proved