# 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 $${45^0}$$ .

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