Question

Verify these identities? sin^2x-tan^2x = -sin^2xtan^2x 1/(secx-tanx) = tanx+secx

Answer 1

Please see below.

Explanation:

.

`sin^2x-tan^2x=sin^2x-sin^2x/cos^2x=sin^2x(1-1/cos^2x)`

But we know:

`1/cos^2x=sec^2x`, therefore:

`sin^2x-tan^2x=sin^2x(1-sec^2x)`

But we have an identity that says:

`sec^2x=1+tan^2x`, therefore,

`sin^2x-tan^2x=sin^2x(1-1-tan^2x)=-sin^2xtan^2x`

`1/(secx-tanx)=(secx+tanx)/((secx-tanx)(secx+tanx))`

`1/(secx-tanx)=(secx+tanx)/(sec^2x-tan^2x`

Using the identity:

`sec^2x=1+tan^2x`, we get:

`1/(secx-tanx)=(secx+tanx)/(1+tan^2x-tan^2x)=secx+tanx`