Question

Prove that √1+tan^2x /√1-sin^2=sec^2x?

Prove that, `sqrt(1+tan^2x)/sqrt(1-sin^2x)=sec^2x.`

Answer 1

Please see below.

Explanation:

We know that,

`color(red)((1)1+tan^2theta=sec^2theta`

`color(blue)((2)1-sin^2theta=cos^2theta`

`color(violet)((3)costheta=1/sectheta`

Here,

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

Let,

`LHS=sqrt(1+tan^2x)/sqrt(1-sin^2x)`

Using `(1) and (2)` we get

`LHS=sqrt(color(red)(sec^2x))/sqrt(color(blue)(cos^2x))`

`color(white)(LHS)=secx/color(violet)(cosx`

`color(white)(LHS)=secx/color(violet)((1/secx)`

`color(white)(LHS)=secx*secx`

`color(white)(LHS)=sec^2x`

`:.LHS=RHS`

If your question is different than above answer,

then type your question again.

Answer 2

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)1+tan^2x=sec^2x`

`rArrsqrt(1+tan^2x)=secx`

`•color(white)(x)sin^2x+cos^2x=1`

`rArrsqrt(1-sin^2x)=cosx`

`"consider the left side"`

`rArrsqrt(1+tan^2x)/sqrt(1-sin^2x)`

`=secx/cosx`

`=1/cosx xx1/cosx`

`=1/cos^2x=sec^2x=" right side "rArr"verified"`