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Prove that : $( sin \theta+ cosec \theta)^{2}+( cos \theta+sec \theta)^{2} =7+ tan^{2} \theta+ cot^{2} \theta$.
Given: $( sin \theta+ cosec \theta)^{2}+( cos \theta+sec \theta)^{2} =7+ tan^{2} \theta+ cot^{2} \theta$.
To do: To prove $L.H.S.=R.H.S.$
Solution:
$L.H.S.=(sin \theta+ cosec \theta)^{2}+(cos \theta+sec \theta)^{2}$
$=sin^{2} \theta+cosec^{2} \theta+2sin \theta cosec \theta+cos^{2} \theta+sec^{2} \theta+2cos \theta sec \theta$
$=sin^{2} \theta+cos^{2} \theta+2sin \theta cosec \theta+sec^{2} \theta+cosec^{2} \theta+2cos \theta sec \theta$
$=1+2+2+sec^{2} \theta+cosec^{2} \theta$ [$\because sin^{2}\theta + cos^{2}\theta=1$, $sin \theta=\frac{1}{cosec \theta}$ and $cos \theta=\frac{1}{sec \theta}$]
$=5+1+tan^{2} \theta+1+cot^{2} \theta$ [$sec^{2} \theta=1+tan^{2} \theta$ and $cosec^{2} \theta=1+cot^{2} \theta$]
$=7+tan^{2} \theta+cot^{2} \theta$
$=R.H.S.$
Hence proved that $L.H.S.=R.H.S.$
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