Question

can anyone help me with this...​

Answer 1

Answer:

L.H.S⇒ {1+cosθ/sinθ+sinθ/cosθ}  (sinθ-cosθ)  /    (1/cos³θ) - (1/sin³θ)

⇒{(sinθcosθ+cos²θ+sin²θ) / (sinθcosθ} (sinθ-cosθ) /

   (sin³θ-cos³θ) / cos³θsin³θ

⇒{(sinθcosθ + 1) / (sinθcosθ)(sinθ-cosθ) /

  (sinθ- cosθ)(sin²θ + sinθcosθ + cos²θ) / sinθcosθ.sin²θcos²θ

⇒(sinθcosθ + 1) (sinθ - cosθ)  /  [( sinθ - cosθ)(sinθcosθ + 1)/sin²θ cos²θ}

⇒sin²θcos²θ

HOPE THAT IT WAS HELPFUL!!!!

MARK IT THE BRAINLIEST IF IT REALLY WAS!!!!!