sin 48^{circ} sec 42^{circ} + cos 48^{circ} o | vedclass

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Question

(A) Given expression: $\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \operatorname{cosec} 42^{\circ}$Using the complementary angle identities: $\s

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given expression: $\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \operatorname{cosec} 42^{\circ}$Using the complementary angle identities: $\sin(90^{\circ} - 	heta) = \cos 	heta$ and $\cos(90^{\circ} - 	heta) = \sin 	heta$.We can write:$\sin 48^{\circ} = \sin(90^{\circ} - 42^{\circ}) = \cos 42^{\circ}$$\cos 48^{\circ} = \cos(90^{\circ} - 42^{\circ}) = \sin 42^{\circ}$Substituting these values into the expression:$= \cos 42^{\circ} \sec 42^{\circ} + \sin 42^{\circ} \operatorname{cosec} 42^{\circ}$Since $\cos 	heta \sec 	heta = 1$ and $\sin 	heta \operatorname{cosec} 	heta = 1$:$= 1 + 1 = 2$

$\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \operatorname{cosec} 42^{\circ} = \ldots \ldots \ldots \ldots$

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