Prove That: Cos15° Cos35° Cosec55° Cos60° Cosec75° = 1 2 - Mathematics

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Sum

Prove that:

cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]

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Solution

\[\begin{array}{l} {\text{LHS}=cos15}^\circ\cos {35}^\circ \cos {ec55}^\circ\cos {60}^\circ \cos {ec75}^\circ \\ \end{array}\]

\[\begin{array}{l}=cos( {90}^0 - {75}^0 )\cos( {90}^0 - {55}^0 )\frac{1}{\sin {55}^0}\times\frac{1}{2}\times\frac{1}{\sin {75}^0} \\ \end{array}\]

\[\begin{array}{l}{=sin75}^0 \sin {55}^0 \frac{1}{\sin {55}^0} \times \frac{1}{2} \times \frac{1}{\sin {75}^0} \\ \end{array}\]\[=\frac{1}{2} = \text{RHS}\]

Concept: Trigonometric Ratios of Complementary Angles
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 6.3 | Page 313
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