
How do you evaluate $ \sec (150) $ ?
Answer
500.7k+ views
Hint: First of all we will find the multiplicative reciprocal or the inverse using sec and cosine relation and then use the identities for trigonometric functions in terms of their complements and refer the trigonometric value table for the resultant required value.
Complete step-by-step answer:
Take the given expression: $ \sec (150) $
By using the multiplicative reciprocal identity for secant and cosine relation:
$ \sec (150) = \dfrac{1}{{\cos (150)}} $
Now, using the trigonometric functions and its complements –
$ \sec (150) = \dfrac{1}{{\cos (150)}} = \dfrac{1}{{\cos (180 - 30)}} $
Since, by using All STC rule cosine is negative in the second quadrant
$ \sec (150) = - \dfrac{1}{{\cos (30)}} $
By referring the trigonometric value table –
$ \sec (150) = - \dfrac{1}{{\dfrac{\sqrt 3}{2}}} $
Denominator’s denominator goes to the numerator.
$ \sec (150) = - \dfrac{{1 \times 2}}{\sqrt 3} $
Find the product of the terms in the numerator. When the denominator is one then we can rewrite it as the number only.
$ \sec (150) = - \dfrac {2}{\sqrt 3}$
This is the required solution.
So, the correct answer is “$ -\dfrac {2}{\sqrt 3}$”.
Note: Remember the trigonometric table having different angles of measures and find the correlation between the six trigonometric functions. Remember the All STC rule, which is also known as the ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ( $ 0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ).
Complete step-by-step answer:
Take the given expression: $ \sec (150) $
By using the multiplicative reciprocal identity for secant and cosine relation:
$ \sec (150) = \dfrac{1}{{\cos (150)}} $
Now, using the trigonometric functions and its complements –
$ \sec (150) = \dfrac{1}{{\cos (150)}} = \dfrac{1}{{\cos (180 - 30)}} $
Since, by using All STC rule cosine is negative in the second quadrant
$ \sec (150) = - \dfrac{1}{{\cos (30)}} $
By referring the trigonometric value table –
$ \sec (150) = - \dfrac{1}{{\dfrac{\sqrt 3}{2}}} $
Denominator’s denominator goes to the numerator.
$ \sec (150) = - \dfrac{{1 \times 2}}{\sqrt 3} $
Find the product of the terms in the numerator. When the denominator is one then we can rewrite it as the number only.
$ \sec (150) = - \dfrac {2}{\sqrt 3}$
This is the required solution.
So, the correct answer is “$ -\dfrac {2}{\sqrt 3}$”.
Note: Remember the trigonometric table having different angles of measures and find the correlation between the six trigonometric functions. Remember the All STC rule, which is also known as the ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ( $ 0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ).
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