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cos theta*cot theta+sin_(1)2theta*csc theta=csc theta

cos⁡θ⋅cot⁡θ+sin⁡12θ⋅csc⁡θ=csc⁡θ \cos \theta \cdot \cot \theta+\sin _{1} 2 \theta \cdot \csc \theta=\csc \theta

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Q. cos⁡θ⋅cot⁡θ+sin⁡12θ⋅csc⁡θ=csc⁡θ \cos \theta \cdot \cot \theta+\sin _{1} 2 \theta \cdot \csc \theta=\csc \theta
  1. Rewrite cot⁡(θ):\cot(\theta): Rewrite cot⁡(θ)\cot(\theta) as cos⁡(θ)sin⁡(θ)\frac{\cos(\theta)}{\sin(\theta)} and csc⁡(θ)\csc(\theta) as 1sin⁡(θ)\frac{1}{\sin(\theta)}.
    \cos(\theta)\cot(\theta) + \sin^\(2(\theta)\csc(\theta) = \cos(\theta)\left(\frac{\cos(\theta)}{\sin(\theta)}\right) + \sin^22(\theta)\left(\frac{11}{\sin(\theta)}\right)
  2. Simplify multiplication: Simplify the expression by performing the multiplication. \newlinecos⁡(θ)(cos⁡(θ)sin⁡(θ))+sin⁡2(θ)(1sin⁡(θ))=cos⁡2(θ)sin⁡(θ)+sin⁡(θ)\cos(\theta)\left(\frac{\cos(\theta)}{\sin(\theta)}\right) + \sin^2(\theta)\left(\frac{1}{\sin(\theta)}\right) = \frac{\cos^2(\theta)}{\sin(\theta)} + \sin(\theta)
  3. Combine terms: Combine the terms over a common denominator. cos⁡2(θ)+sin⁡2(θ)sin⁡(θ)\frac{\cos^2(\theta) + \sin^2(\theta)}{\sin(\theta)}
  4. Use Pythagorean identity: Use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1. (1)/sin⁡(θ)(1)/\sin(\theta)
  5. Recognize csc⁡(θ)\csc(\theta): Recognize that 1sin⁡(θ)\frac{1}{\sin(\theta)} is csc⁡(θ)\csc(\theta).csc⁡(θ)\csc(\theta)

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