
What is $\left( {\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{\cos ec18^\circ }}{{\cos ec144^\circ }}} \right)$ equals to?
A. $\sec 18^\circ $
B.$\cos ec18^\circ $
C.-$\sec 18^\circ $
D.-$\cos ec18^\circ $
Answer
581.7k+ views
Hint: To find the value of the above expression we will break the 144 as 180 – 36. Then use the transformation of trigonometric ratio formula and simplify such that we will get the value of this expression.
Complete step-by-step answer:
First write 144 = 180 - 36 and proceed we will get,
$
= \left( {\dfrac{{\sec 18^\circ }}{{\sec \left( {180^\circ - 36^\circ } \right)}} + \dfrac{{\cos ec18^\circ }}{{\cos ec\left( {180^\circ - 36^\circ } \right)}}} \right) \\
= \left( {\dfrac{{\sec 18^\circ }}{{ - sec36^\circ }} + \dfrac{{\cos ec18^\circ }}{{\cos ec36^\circ }}} \right) \\
$
Now converting the above expression in the form of sin and cos
We will get,
$
= \left( { - \dfrac{{\dfrac{1}{{\cos 18^\circ }}}}{{\dfrac{1}{{\cos 36^\circ }}}} + \dfrac{{\dfrac{1}{{\sin 18^\circ }}}}{{\dfrac{1}{{\sin 36^\circ }}}}} \right) \\
= \left( { - \dfrac{{\cos 36^\circ }}{{\cos 18^\circ }} + \dfrac{{\sin 36^\circ }}{{\sin 18^\circ }}} \right) \\
$
On taking LCM and solving we get,
$ = \left( {\dfrac{{ - \cos 36^\circ \sin 18^\circ + \sin 36^\circ \cos 18^\circ }}{{\cos 18^\circ \sin 18^\circ }}} \right)$
Use the sin (A-B) formula then simplify we get,
$
= \left( {\dfrac{{\sin \left( {36^\circ - 18^\circ } \right)}}{{\cos 18^\circ \sin 18^\circ }}} \right) \\
= \left( {\dfrac{{\sin 18^\circ }}{{\cos 18^\circ \sin 18^\circ }}} \right) \\
= \dfrac{1}{{\cos 18^\circ }} \\
= \sec 18^\circ \\
$
The above expression is equals to $\sec 18^\circ $
So, the correct answer is “Option A”.
Note: Scientific calculators have sin, cos, and tan functions, as well as the inverse functions. It's worth taking a few minutes to work out.
Trigonometric ratios sin and cosec are positive in the 1st and 2nd quadrant and in the 3rd and 4th quadrant are negative. Cos and sec are positive in the 1st and 4th quadrant. Tan and cot are positive in the 3rd and 1st quadrant.
Complete step-by-step answer:
First write 144 = 180 - 36 and proceed we will get,
$
= \left( {\dfrac{{\sec 18^\circ }}{{\sec \left( {180^\circ - 36^\circ } \right)}} + \dfrac{{\cos ec18^\circ }}{{\cos ec\left( {180^\circ - 36^\circ } \right)}}} \right) \\
= \left( {\dfrac{{\sec 18^\circ }}{{ - sec36^\circ }} + \dfrac{{\cos ec18^\circ }}{{\cos ec36^\circ }}} \right) \\
$
Now converting the above expression in the form of sin and cos
We will get,
$
= \left( { - \dfrac{{\dfrac{1}{{\cos 18^\circ }}}}{{\dfrac{1}{{\cos 36^\circ }}}} + \dfrac{{\dfrac{1}{{\sin 18^\circ }}}}{{\dfrac{1}{{\sin 36^\circ }}}}} \right) \\
= \left( { - \dfrac{{\cos 36^\circ }}{{\cos 18^\circ }} + \dfrac{{\sin 36^\circ }}{{\sin 18^\circ }}} \right) \\
$
On taking LCM and solving we get,
$ = \left( {\dfrac{{ - \cos 36^\circ \sin 18^\circ + \sin 36^\circ \cos 18^\circ }}{{\cos 18^\circ \sin 18^\circ }}} \right)$
Use the sin (A-B) formula then simplify we get,
$
= \left( {\dfrac{{\sin \left( {36^\circ - 18^\circ } \right)}}{{\cos 18^\circ \sin 18^\circ }}} \right) \\
= \left( {\dfrac{{\sin 18^\circ }}{{\cos 18^\circ \sin 18^\circ }}} \right) \\
= \dfrac{1}{{\cos 18^\circ }} \\
= \sec 18^\circ \\
$
The above expression is equals to $\sec 18^\circ $
So, the correct answer is “Option A”.
Note: Scientific calculators have sin, cos, and tan functions, as well as the inverse functions. It's worth taking a few minutes to work out.
Trigonometric ratios sin and cosec are positive in the 1st and 2nd quadrant and in the 3rd and 4th quadrant are negative. Cos and sec are positive in the 1st and 4th quadrant. Tan and cot are positive in the 3rd and 1st quadrant.
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