prove that cosec0/(tan0+cot0)=cos0
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Answer:
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here is your answer
Consider the provided expression.
(\csc\theta-\sin \theta)(\sec \theta - \cos \theta)=\dfrac{1}{\tan \theta + \cot\theta}
Consider the LHS.
=(\csc\theta-\sin \theta)(\sec \theta - \cos \theta)\\=(\frac{1}{\sin\theta}-\sin \theta)(\frac{1}{\cos\theta} - \cos \theta)\\=(\frac{1-\sin^2\theta}{\sin\theta})(\frac{1- \cos^2 \theta}{\cos\theta})\\=(\frac{\cos^2\theta}{\sin\theta})(\frac{\sin^2 \theta}{\cos\theta})\\=\cos\theta\sin\theta
Now Consider the RHS
=\dfrac{1}{\tan \theta + \cot\theta}\\=\dfrac{1}{\frac{\sin\theta}{\cos\theta} +\frac{\cos\theta}{\sin\theta}}\\\\=\dfrac{\cos\theta\sin\theta}{{\sin^2\theta}+{\cos^2\theta}}\\\\=\cos\theta\sin\theta
LHS=RHS
Hence, proved
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