Question #895e3

1 Answer
Sep 8, 2015

For solving this problem we need to know some trigonometric ratios and identities:
1)cosectheta=1/sintheta
2)cottheta=cos theta/sintheta
3)sin^2theta+cos^2theta=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 (cosectheta- cottheta)^2

2) using 1 & 2 rArr (1/sintheta-costheta/sintheta)^2

3) now we get rArr (1-costheta/sintheta)^2

4) on splitting we get rArr (1-costheta)^2 / (sintheta^2)

5)now (1-costheta)^2 can be written as :
(1-costheta) * (1-costheta) and (sintheta^2) can be written as :

(1-costheta^2) = (1-costheta) * (1+costheta)#

6) so,we get
(1-costheta) * (1-costheta) / (1-costheta) * (1+costheta)

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