Without using trigonometric tables,evaluate: | vedclass

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(B) We know the complementary angle identities: $\sin(90^{\circ} - 	heta) = \cos 	heta$ and $\cos(90^{\circ} - 	heta) = \sin 	heta$.Also,$\sec 	h

Vedclass · Accepted Answer

$2$

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(B) We know the complementary angle identities: $\sin(90^{\circ} - 	heta) = \cos 	heta$ and $\cos(90^{\circ} - 	heta) = \sin 	heta$.Also,$\sec 	heta = \frac{1}{\cos 	heta}$ and $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta}$.Given expression: $\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \operatorname{cosec} 42^{\circ}$.Substitute $48^{\circ} = 90^{\circ} - 42^{\circ}$:$= \sin(90^{\circ} - 42^{\circ}) \sec 42^{\circ} + \cos(90^{\circ} - 42^{\circ}) \operatorname{cosec} 42^{\circ}$$= \cos 42^{\circ} \sec 42^{\circ} + \sin 42^{\circ} \operatorname{cosec} 42^{\circ}$$= \cos 42^{\circ} \cdot \frac{1}{\cos 42^{\circ}} + \sin 42^{\circ} \cdot \frac{1}{\sin 42^{\circ}}$$= 1 + 1 = 2$.

Without using trigonometric tables,evaluate: $\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \operatorname{cosec} 42^{\circ} = $

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