Question

Given that a = sec x + cosec x and b = sec x - cosec x, show that a^2 + b^2 = 2 sec^2 x . cosec^2 x. Please?

I got 2(sec^2 x + cosec^2 x) instead of it. Is the question wrong or something else? It is an IGCSE past paper question (Additional Math)

Answer 1

`"see explanation"`

Explanation:

`"note that "(a+b)^2=a^2+2ab+b^2`

`rArra^2+b^2=(a+b)^2-2ab`

`(a+b)^2=(secx+cscx+secx-cscx)^2=(2secx)^2=4sec^2x`

`ab=(secx-cscx)(secx-cscx)=sec^2x-csc^2x`

`rArra^2+b^2-2ab`

`=4sec^2x-2sec^2x+2csc^2x=2sec^2x+2csc^2x`

`=2/cos^2x+2/sin^2x`

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

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

`=2/(cos^2xsin^2x)`

`=2xx1/(cos^2x)xx1/sin^2x`

`=2sec^2xcsc^2x`

Answer 2

Please see below.

Explanation:

As `a=secx+cscx` and `b=secx-cscx`

Hence `a^2+b^2`

= `(secx+cscx)^2+(secx-cscx)^2`

= `sec^2x+csc^2x+2secxcscx+sec^2x+csc^2x-2secxcscx`

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

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

= `2((sin^2x+cos^2x)/(sin^2xcos^2x))`

= `2/(sin^2xcos^2x)`

= `2sec^2xcsc^2x`