tan theta + cot theta = ldots ldots ldots | vedclass

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(D) We know that $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$ and $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$.Substituting these into the expre

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$\operatorname{cosec} 	heta \cdot \sec 	heta$

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(D) We know that $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$ and $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$.Substituting these into the expression: $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta}$.Taking the common denominator: $\frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta}$.Using the trigonometric identity $\sin^2 	heta + \cos^2 	heta = 1$,we get: $\frac{1}{\sin 	heta \cos 	heta}$.Since $\frac{1}{\sin 	heta} = \operatorname{cosec} 	heta$ and $\frac{1}{\cos 	heta} = \sec 	heta$,the expression simplifies to $\operatorname{cosec} 	heta \cdot \sec 	heta$.

$\tan \theta + \cot \theta = \ldots \ldots \ldots$

Similar Questions

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$(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2} = \dots$

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