Question

prove that -1+sinAsin(90°-A)/cot(90°-A)=-sin^2A

Answer 1

Ur answer is in the pic


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Answer 2

$\huge\fcolorbox{black}{teal}{Solution:-}$

✏️ To prove:-

→ -1 + sinA sin (90° - A)/ cot (90° - A) = - sin²A

✏️ Proof:-

→ LHS = -1 + sinA sin(90°-A)/cot (90°-A)

[∵ sin (90°- θ) = cosθ]

[∵ cot (90°- θ) = tanθ]

= -1 + sinAcosA/tanA

= -1 + sinAcosA × cotA

[∵ cotθ = cosθ/sinθ]

= -1 + sinAcosA × cosA/sinA

[∵ sin²θ + cos²θ = 1]

= -1 + cos²A = -(1-cos²A)

= -sin²A = RHS

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I hope this helps! :)