Solveeit Logo
Login

Question

Question: What is the value of \(\cot {{60}^{\circ }}-\sec {{45}^{\circ }}\)? (a) \(\dfrac{\sqrt{2}-\sqrt{3}...

What is the value of cot⁡60∘−sec⁡45∘\cot {{60}^{\circ }}-\sec {{45}^{\circ }}?
(a) 2−36\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}}
(b) 3−323\dfrac{\sqrt{3}-3\sqrt{2}}{3}
(c) 1−222\dfrac{1-2\sqrt{2}}{2}
(d) 1−32\dfrac{1-\sqrt{3}}{2}

Explanation

Solution

Hint: Convert co-tangent of the given angle into cos⁡θsin⁡θ\dfrac{\cos \theta }{\sin \theta } and secant of the given angle into 1cos⁡θ\dfrac{1}{\cos \theta } and put the required value of angle. Then rationalize the denominator, if required, to get the answer.

Complete step-by-step answer:

Trigonometric values of different ratios, such as sine, cosine, tangent, secant, cosecant and cotangent deal with the measurement of lengths and angles of any right angle triangle. The values of trigonometric functions for 0∘,30∘,45∘,60∘ and 90∘{{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}\text{ and }{{90}^{\circ }} are commonly used to solve the question. Trigonometric functions values are all about the study of standard angles for a given triangle with respect to trigonometric ratios. Trigonometry is one of the most important parts of geometry where the relationship between angle and sides of a triangle are discussed.
We know that cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } and\text{and} sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta }. Also, it is already known that, sin⁡60∘=32, cos⁡60∘=12 and cos⁡45∘=12\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2},\text{ }\cos {{60}^{\circ }}=\dfrac{1}{2}\text{ and }\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}.
Now, let us come to the question. We have to find the value of cot⁡60∘−sec⁡45∘\cot {{60}^{\circ }}-\sec {{45}^{\circ }}.
cot⁡60∘−sec⁡45∘ =cos⁡60∘sin⁡60∘−1cos⁡45∘ =1232−112 =13−2 =1−2×33 =1−63 \begin{aligned} & \cot {{60}^{\circ }}-\sec {{45}^{\circ }} \\\ & =\dfrac{\cos {{60}^{\circ }}}{\sin {{60}^{\circ }}}-\dfrac{1}{\cos {{45}^{\circ }}} \\\ & =\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}-\dfrac{1}{\dfrac{1}{\sqrt{2}}} \\\ & =\dfrac{1}{\sqrt{3}}-\sqrt{2} \\\ & =\dfrac{1-\sqrt{2}\times \sqrt{3}}{\sqrt{3}} \\\ & =\dfrac{1-\sqrt{6}}{\sqrt{3}} \\\ \end{aligned}
Now, rationalizing the denominator in the obtained expression, we get,
cot⁡60∘−sec⁡45∘ =1−63×33 =3−183×3 =3−323 \begin{aligned} & \cot {{60}^{\circ }}-\sec {{45}^{\circ }} \\\ & =\dfrac{1-\sqrt{6}}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}} \\\ & =\dfrac{\sqrt{3}-\sqrt{18}}{\sqrt{3}\times \sqrt{3}} \\\ & =\dfrac{\sqrt{3}-3\sqrt{2}}{3} \\\ \end{aligned}
Hence, option (b) is the correct answer.

Note: As you can see that first we have converted cot⁡θ\cot \theta into cos⁡θsin⁡θ\dfrac{\cos \theta }{\sin \theta } and sec⁡θ\sec \theta into 1cos⁡θ\dfrac{1}{\cos \theta } and then used their particular values for the given angle. But, we can directly put the values, cot⁡60∘=13\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}} and sec⁡45∘=2\sec {{45}^{\circ }}=\sqrt{2} then simplify and rationalize the expression. This will save our time and steps. To do this we need to memorize the values.