Trig1.3

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Prove that $\cot^2 \theta[\frac{\sec\theta-1}{1+\sin\theta}]+\sec^2 \theta[\frac{\sin\theta-1}{1+\sec\theta}]=0\,$

L.H.S= $\cot^2 \theta[\frac{\sec\theta-1}{1+\sin\theta}]+\sec^2 \theta[\frac{\sin\theta-1}{1+\sec\theta}]\,$

$\frac{\cot^2\theta(\sec^2\theta-1)+\sec^2\theta(\sin^2\theta-1)}{(1+\sin\theta)(1+\sec\theta)}\,$

$\frac{\cot^2\theta(\tan^2\theta)+\sec^2\theta(-\cos^2\theta)}{(1+\sin\theta)(1+\sec\theta)}=\frac{1-1}{(1+\sin\theta)(1+\sec\theta)}=0\,$ =RHS

Main Page:Trigonometry

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