Question

plz solve this its needed urgently

Answer 1

$\sf{cosec \theta - sin \theta = x}$


${\boxed{\sf{cosec \theta = {\dfrac{1}{sin \theta}}}}}$


$\therefore$


$\sf{{\dfrac{1}{sin \theta}} - sin \theta = x}$


$\sf{{\dfrac{1 - sin^2 \theta}{sin \theta}} = x}$


${\boxed{\sf{1 - sin^2 \theta = cos^2 \theta}}}$


$\sf{{\dfrac{cos^2 \theta}{sin \theta}} = x}$


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$\sf{tan \theta + cot \theta = y}$


${\boxed{\sf{cot \theta = {\dfrac{1}{tan \theta}}}}}$


$\therefore$


$\sf{tan \theta + {\dfrac{1}{tan \theta}} = y}$


$\sf{{\dfrac{tan^2 \theta + 1}{tan \theta}} = y}$


${\boxed{\sf{tan^2 \theta + 1 = sec^2 \theta}}}$


$\sf{{\dfrac{sec^2 \theta}{tan \theta}} = y}$


${\boxed{\sf{sec^2 \theta = {\dfrac{1}{cos^2 \theta}}}}}$


${\boxed{\sf{tan \theta = {\dfrac{sin \theta}{cos \theta}}}}}$


$\therefore$


$\sf{ {\dfrac{ {\dfrac{1}{cos^2 \theta}} }{ {\dfrac{sin \theta}{cos \theta}} }} = y}$


$\sf{ {\dfrac{1}{cos^2 \theta}} × {\dfrac{cos \theta}{sin \theta}} = y}$


$\sf{{\dfrac{1}{cos \theta . sin \theta}} = y}$


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See the Attachment for the further solution.