Answer :
Given
⇒x cot (90° + θ) + tan (90° + θ) sin θ + cosec (90° + θ) = 0
⇒x [-tan θ] + [-cot θ] sin θ + sec θ = 0
⇒ -x sin θ – cos2 θ + 1 = 0
We know that 1 – cos2 θ = sin2 θ
⇒ -x sin θ + sin2 θ = 0
⇒x sin θ = sin2 θ
⇒x = sin2 θ/ sin θ
∴x = sin θ
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