Question

How do you prove `(1-sin^2theta)(1+cot^2theta)=cot^2theta`?

Answer 1

Please see below.

Explanation:

We know that ,

`(1)cos^2x+sin^2x=1`
`(2)csc^2x-cot^2x=1`
`(3)cscx=1/sinx`
`(4)cosx/sinx=cotx`

Using `(1) and (2):`

`LHS=(1-sin^2theta)(1+cot^2theta)`

`LHS=cos^2thetacsc^2thetatoApply(3)`

`LHS=cos^2theta*1/sin^2theta`

`LHS=cos^2theta/sin^2thetatoApply(4)`

`LHS=cot^2theta`

`LHS=RHS`

Answer 2

`"see explanation"`

Explanation:

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

`•color(white)(x)cottheta=costheta/sintheta`

`"consider the left side"`

`(1-sin^2theta)(1+cos^2theta/sin^2theta)`

`"expand the factors"`

`=1+cot^2theta-sin^2theta-cos^2theta`

`=1+cot^2theta-(sin^2theta+cos^2theta)`

`=1+cot^2theta-1larrsin^2theta+cos^2theta=1`

`=cot^2theta=" right side "rArr" verified"`

Answer 3

As proved below.

Explanation:

![https://in.pinterest.com/pin/359021401516045145/]()

`sin^2 theta + cos^2 theta = 1`

`1 - sin^2 theta = cos^2 theta`

`1 + cot^2 theta = csc^2 theta`

`csc theta = 1/ sin theta`

`(1 - sin^2 theta) (1 + cot^2 theta) = cos^2 theta * csc^2 theta`

`=> cos^2 theta * 1 / sin^2 theta`

`=> cos^2 theta / sin^2 theta = cot ^2 theta = R H S`