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प्रश्न
Prove that:
cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]
बेरीज
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उत्तर
\[\begin{array}{l} {\text{LHS}=cos15}^\circ\cos {35}^\circ \cos {ec55}^\circ\cos {60}^\circ \cos {ec75}^\circ \\ \end{array}\]
\[\begin{array}{l}=cos( {90}^0 - {75}^0 )\cos( {90}^0 - {55}^0 )\frac{1}{\sin {55}^0}\times\frac{1}{2}\times\frac{1}{\sin {75}^0} \\ \end{array}\]
\[\begin{array}{l}{=sin75}^0 \sin {55}^0 \frac{1}{\sin {55}^0} \times \frac{1}{2} \times \frac{1}{\sin {75}^0} \\ \end{array}\]\[=\frac{1}{2} = \text{RHS}\]
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