tan (65^circ - theta) - cot (25^circ + theta) | vedclass

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(D) We know the trigonometric identities for complementary angles: $	an(90^\circ - A) = \cot A$ and $\sec(90^\circ - A) = \operatorname{cosec} A$.Fir

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(D) We know the trigonometric identities for complementary angles: $	an(90^\circ - A) = \cot A$ and $\sec(90^\circ - A) = \operatorname{cosec} A$.First,consider the term $	an(65^\circ - 	heta)$:$	an(65^\circ - 	heta) = \cot(90^\circ - (65^\circ - 	heta)) = \cot(25^\circ + 	heta)$.Next,consider the term $\sec(55^\circ - 	heta)$:$\sec(55^\circ - 	heta) = \operatorname{cosec}(90^\circ - (55^\circ - 	heta)) = \operatorname{cosec}(35^\circ + 	heta)$.Now,substitute these into the given expression:$	an(65^\circ - 	heta) - \cot(25^\circ + 	heta) - \sec(55^\circ - 	heta) + \operatorname{cosec}(35^\circ + 	heta)$$= \cot(25^\circ + 	heta) - \cot(25^\circ + 	heta) - \operatorname{cosec}(35^\circ + 	heta) + \operatorname{cosec}(35^\circ + 	heta)$$= 0$.

$\tan (65^\circ - \theta) - \cot (25^\circ + \theta) - \sec (55^\circ - \theta) + \operatorname{cosec}(35^\circ + \theta) = \ldots \ldots \ldots \ldots$ (where,$0 < \theta < 25^\circ$)

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