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CBSE Class 10 Math 2023 All Sets Solved Paper

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Question : 4 of 20
Marks: +1, -0
  cos⁡2θsin⁡2θ−  1sin⁡2θ,\;\frac{\cos^2 θ}{\sin^2 θ} - \;\frac{1}{\sin^2 θ}, in simplified form, is
Solution:  
  cos⁡2θsin⁡2θ−  1sin⁡2θ\;\frac{\cos^2 θ}{\sin^2 θ} - \;\frac{1}{\sin^2 θ}
=  cos⁡2−1sin⁡2θ=\;\frac{\cos^2 -1}{\sin^2 θ}
=  −(1−cos⁡2θ)sin⁡2θ=\;\frac{-(1- \cos^2 θ)}{\sin^2 θ} [∵sin⁡2θ+cos⁡2θ=1, sin⁡2θ=1−cos⁡2θ][ \because \sin^2 θ + \cos^2 θ = 1,\ \sin^2 θ =1-\cos^2 θ]
=−  sin⁡2θsin⁡2θ=−1=-\;\frac{\sin^2 θ}{\sin^2 θ} = -1
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