cot (45^circ + theta ) cot (45^circ - theta ) | vedclass

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(C) We know that $\cot(A) = 	an(90^\circ - A)$.Applying this to the first term: $\cot(45^\circ + 	heta) = 	an(90^\circ - (45^\circ + 	heta)) = 	a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) We know that $\cot(A) = 	an(90^\circ - A)$.Applying this to the first term: $\cot(45^\circ + 	heta) = 	an(90^\circ - (45^\circ + 	heta)) = 	an(45^\circ - 	heta)$.Substituting this back into the expression:$\cot(45^\circ + 	heta) \cot(45^\circ - 	heta) = 	an(45^\circ - 	heta) \cot(45^\circ - 	heta)$.Since $	an(x) \cot(x) = 1$ for any angle $x$,we have:$	an(45^\circ - 	heta) \cot(45^\circ - 	heta) = 1$.

$\cot (45^\circ + \theta ) \cot (45^\circ - \theta ) = $

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