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Prove the following trigonometric identities:\(\frac{cos^2θ}{sin θ}-cosec θ+sin θ=0\) |
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Answer» \(\frac{cos^2θ}{sin θ}-cosec θ+sin θ \) = \(\frac{cos^2θ} {sin θ}-\frac{1}{sin θ}+sinθ\) = \(\frac{cos^2θ-1+sin^2θ} {sin θ}\) = \(\frac{(cos^2θ+sin^2θ)-1}{sinθ}\) = \(\frac{1-1}{sin θ}\) =0 Hence Proved. |
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