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Express $sin\ 67^o + cos\ 75^o$ in terms of trigonometric ratios of angles between $0^o$ and $45^o$.
Given:
\( \sin 67^{\circ}+\cos 75^{\circ} \)
To do:
We have to express \( \sin 67^{\circ}+\cos 75^{\circ} \) in terms of trigonometric ratios of angles lying between \( 0^{\circ} \) and \( 45^{\circ} \).
Solution:
We know that,
$cos\ (90^{\circ}- \theta) = sin\ \theta$
$sin\ (90^{\circ}- \theta) = cos\ \theta$
Therefore,
$\sin 67^{\circ}+\cos 75^{\circ}=\sin (90^{\circ}-23^{\circ})+\cos (90^{\circ}-15^{\circ})$
$=\cos 23^{\circ}+\sin 15^{\circ}$
Therefore, $\sin 67^{\circ}+\cos 75^{\circ}=\cos 23^{\circ}+\sin 15^{\circ}$.
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