Question

Question #0860e

Answer 1

Refer to The Explanation.

Explanation:

Recall that, `sin^2x+cos^2x=1, and, tan^2x=1/cot^2x.`

Hence,`" the L.H.S.="(1+cot^2x)/{1+1/cot^2x},`

`=(1+cot^2x)/{(1+cot^2x)/cot^2x},`

`=(cancel(1+cot^2x)){cot^2x/(cancel(1+cot^2x))},`

`=cot^2x,`

`"=the R.H.S."`

Answer 2

`"LS"=frac{sin^2x+cos^2x+cot^2x}{1+tan^2x}`

`=frac{1+cot^2x}{1+tan^2x}`

Since `color(red)(sin^2a+cos^2a=1),` (Pythagorean identity) dividing by `sin^2a` gives another identity `color(red)(1+cot^2a=csc^2a=1/sin^2a)`, and dividing the Pythagorean identity by `cos^2a` gives `color(red)(tan^2a+1=sec^2a=1/cos^2a)`.

`"LS"=frac{(1/sin^2x)}{(1/cos^2x)}`

`=frac{cos^2x}{sin^2x}`

`=cot^2x`

`"QED"`