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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Updated on: 10-Oct-2022

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