Ex.8.3 Q7 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

Express \(\sin {67^0} + \cos {75^0}\) in terms of trigonometric ratios of angles between \({0^0}\) and \({45^0}\) .

Text Solution

 

Reasoning:

\(\cos \left( {{{90}^0} - \theta } \right) = \sin \theta \)

Steps:

Given that: \(\sin {67^0} + \cos {75^0}\) ….(i)

Since \(\cos \left( {{{90}^0} - \theta } \right) = \sin \theta \)

By using property in equation (i) we get:

\[\begin{align} &= \sin \left( {{{90}^0} - {{23}^0}} \right) + \cos \left( {{{90}^0} - {{15}^0}} \right)\\ &= \cos {23^0} + \sin {15^0} \end{align}\]

Hence, the expression \(\cos {23^0} + \sin {15^0}\) has trigonometric ratios of angles between \({0^0}\) and \({45^0}\) .