Question

Question #895e3

Answer 1

For solving this problem we need to know some trigonometric ratios and identities:
1)cosec`theta`=1/sin`theta`
2)cot`theta`=cos `theta`/sin`theta`
3)`sin^2` `theta+cos^2` `theta=1`

Explanation:

For an easy solution to all proof questions its convenient to go for
L.H.S= R.H.S method:

1)in R.H.S (WE HAVE) `rArr` `(cosec` `theta`- `cot` `theta` `)^2`

2) using 1 & 2 `rArr` `(1/sin` `theta`-cos`theta`/`sin` `theta` `)^2`

3) now we get `rArr` `(1-cos` `theta`/`sin` `theta` `)^2`

4) on splitting we get `rArr` (`1-cos` `theta)^2` / (`sin` `theta^2`)

5)now (`1-cos` `theta)^2` can be written as :
(`1-cos` `theta`) * `(1-cos` `theta`)# and (#sin`theta^2`) can be written as :

(`1-cos` `theta^2`) = (`1-cos` `theta`) * `(1+cos` `theta`)#

6) so,we get
(`1-cos` `theta`) * (`1-cos` `theta`) / (`1-cos` `theta`) * (`1+cos` `theta`)

7) Finally on cancelling we get ,
(`1-cos` `theta`) / (`1+cos` `theta`).......(so L.H.S= R.H.S)........... (Proved)
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I hope this helps :)