sec 55^{circ} cdot sin 35^{circ} + cos 35^{ci | vedclass

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(D) We know that $\sec 	heta = \frac{1}{\cos 	heta}$ and $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta}$.Also,$\sin(90^{\circ} - 	heta) = \co

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$2$

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(D) We know that $\sec 	heta = \frac{1}{\cos 	heta}$ and $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta}$.Also,$\sin(90^{\circ} - 	heta) = \cos 	heta$ and $\cos(90^{\circ} - 	heta) = \sin 	heta$.Given expression: $\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ}$.Convert $55^{\circ}$ to $(90^{\circ} - 35^{\circ})$:$= \sec(90^{\circ} - 35^{\circ}) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec}(90^{\circ} - 35^{\circ})$$= \operatorname{cosec} 35^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \sec 35^{\circ}$Since $\operatorname{cosec} 35^{\circ} = \frac{1}{\sin 35^{\circ}}$ and $\sec 35^{\circ} = \frac{1}{\cos 35^{\circ}}$:$= \left(\frac{1}{\sin 35^{\circ}}ight) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \left(\frac{1}{\cos 35^{\circ}}ight)$$= 1 + 1 = 2$.

$\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ} = \ldots \ldots \ldots \ldots$

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