Question #895e3

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Question #895e3

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Answer 1 · 2015-09-08T19:28:55

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

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) $\Rightarrow$ $rArr$ $(cos e c$ $(cosec$ $θ$ $theta$ - $cot$ $cot$ $θ$ $theta$ $)^{2}$ $)^2$

2) using 1 & 2 $\Rightarrow$ $rArr$ $(\frac{1}{sin}$ $(1/sin$ $θ$ $theta$ -cos $θ$ $theta$ / $sin$ $sin$ $θ$ $theta$ $)^{2}$ $)^2$

3) now we get $\Rightarrow$ $rArr$ $(1 - cos$ $(1-cos$ $θ$ $theta$ / $sin$ $sin$ $θ$ $theta$ $)^{2}$ $)^2$

4) on splitting we get $\Rightarrow$ $rArr$ ( $1 - cos$ $1-cos$ $θ)^{2}$ $theta)^2$ / ( $sin$ $sin$ $θ^{2}$ $theta^2$ )

5)now ( $1 - cos$ $1-cos$ $θ)^{2}$ $theta)^2$ can be written as :
( $1 - cos$ $1-cos$ $θ$ $theta$ ) * $(1 - cos$ $(1-cos$ $θ$ $theta$ ) $and ($ $and ($ sin $θ^{2}$ $theta^2$ ) can be written as :

( $1 - cos$ $1-cos$ $θ^{2}$ $theta^2$ ) = ( $1 - cos$ $1-cos$ $θ$ $theta$ ) * $(1 + cos$ $(1+cos$ $θ$ $theta$ )#

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

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

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Question #895e3

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