What is the value of $\cot {75^ \circ }$ ?
Answer
545.4k+ views
Hint: Here we are going to find the value of $\cot {75^ \circ }$by using trigonometry property and also the formula.
Formula used:
We know that $\cot x = \dfrac{1}{{\tan x}}$ and
The trigonometry identity $\tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 - \tan A.\tan B}}$
Complete step-by-step solution:
Using the trigonometry identity that is $\tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 - \tan A.\tan B}}$------------(1)
And we have $\cot {75^ \circ }$and we can split the angle like this $\cot \left( {{{30}^ \circ } + {{45}^ \circ }} \right)$
We know $\cot x = \dfrac{1}{{\tan x}}$so we can write the above like this $\cot \left( {{{30}^ \circ } + {{45}^ \circ }} \right) = \dfrac{1}{{\tan \left( {{{30}^ \circ } + {{45}^ \circ }} \right)}}$
Now we can find the value for $\tan \left( {{{30}^ \circ } + {{45}^ \circ }} \right)$by using trigonometry identity we get,
$\tan \left( {{{30}^ \circ } + {{45}^ \circ }} \right) = \dfrac{{\tan {{30}^ \circ } + \tan {{45}^ \circ }}}{{1 - \tan {{30}^ \circ }.\tan {{45}^ \circ }}}$-------------(2)
We know the value that $\tan {45^ \circ } = 1$, $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
Substituting these value in equation(1) we get,
$\tan \left( {{{75}^ \circ }} \right) = \dfrac{{\dfrac{1}{{\sqrt 3 }} + 1}}{{1 - \dfrac{1}{{\sqrt 3 }}\left( 1 \right)}}$
On simplifying it we get,
$\tan \left( {{{75}^ \circ }} \right) = \dfrac{{\dfrac{{1 + \sqrt 3 }}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}$
$\tan \left( {{{75}^ \circ }} \right) = \dfrac{{1 + \sqrt 3 }}{{\sqrt 3 - 1}}$ ----------(3)
Using the identity $\cot x = \dfrac{1}{{\tan x}}$we get,
$\cot {75^ \circ } = \dfrac{1}{{\tan {{75}^ \circ }}}$
Therefore substituting equation (3) in above equation we get,
$\cot {75^ \circ } = \dfrac{1}{{\dfrac{{1 + \sqrt 3 }}{{\sqrt 3 - 1}}}}$
On simplifying we get,
$\cot {75^ \circ } = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }}$
Finally we get the answer.
Note: In a right triangle, the cotangent of an angle is the length of the adjacent side divided by the length of the opposite side. In a formula, it is abbreviated to just ‘cot’ . Trigonometry values are all about the study of standard angles for a given triangle with respect to trigonometric ratios. The word ‘trigon’ means triangle and ‘ metron’ means measurement. It’s one of the major concepts and part of geometry ,where the relationship between angles and sides of a triangle is explained.
Like other trigonometric functions, the cotangent can be represented as a line segment associated with the unit circle. Obviously , since the cotangent function is the reciprocal of the tangent function , it can be expressed in terms of tangent. We can also express the cotangent function in terms of the sine and cosine.
Formula used:
We know that $\cot x = \dfrac{1}{{\tan x}}$ and
The trigonometry identity $\tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 - \tan A.\tan B}}$
Complete step-by-step solution:
Using the trigonometry identity that is $\tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 - \tan A.\tan B}}$------------(1)
And we have $\cot {75^ \circ }$and we can split the angle like this $\cot \left( {{{30}^ \circ } + {{45}^ \circ }} \right)$
We know $\cot x = \dfrac{1}{{\tan x}}$so we can write the above like this $\cot \left( {{{30}^ \circ } + {{45}^ \circ }} \right) = \dfrac{1}{{\tan \left( {{{30}^ \circ } + {{45}^ \circ }} \right)}}$
Now we can find the value for $\tan \left( {{{30}^ \circ } + {{45}^ \circ }} \right)$by using trigonometry identity we get,
$\tan \left( {{{30}^ \circ } + {{45}^ \circ }} \right) = \dfrac{{\tan {{30}^ \circ } + \tan {{45}^ \circ }}}{{1 - \tan {{30}^ \circ }.\tan {{45}^ \circ }}}$-------------(2)
We know the value that $\tan {45^ \circ } = 1$, $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
Substituting these value in equation(1) we get,
$\tan \left( {{{75}^ \circ }} \right) = \dfrac{{\dfrac{1}{{\sqrt 3 }} + 1}}{{1 - \dfrac{1}{{\sqrt 3 }}\left( 1 \right)}}$
On simplifying it we get,
$\tan \left( {{{75}^ \circ }} \right) = \dfrac{{\dfrac{{1 + \sqrt 3 }}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}$
$\tan \left( {{{75}^ \circ }} \right) = \dfrac{{1 + \sqrt 3 }}{{\sqrt 3 - 1}}$ ----------(3)
Using the identity $\cot x = \dfrac{1}{{\tan x}}$we get,
$\cot {75^ \circ } = \dfrac{1}{{\tan {{75}^ \circ }}}$
Therefore substituting equation (3) in above equation we get,
$\cot {75^ \circ } = \dfrac{1}{{\dfrac{{1 + \sqrt 3 }}{{\sqrt 3 - 1}}}}$
On simplifying we get,
$\cot {75^ \circ } = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }}$
Finally we get the answer.
Note: In a right triangle, the cotangent of an angle is the length of the adjacent side divided by the length of the opposite side. In a formula, it is abbreviated to just ‘cot’ . Trigonometry values are all about the study of standard angles for a given triangle with respect to trigonometric ratios. The word ‘trigon’ means triangle and ‘ metron’ means measurement. It’s one of the major concepts and part of geometry ,where the relationship between angles and sides of a triangle is explained.
Like other trigonometric functions, the cotangent can be represented as a line segment associated with the unit circle. Obviously , since the cotangent function is the reciprocal of the tangent function , it can be expressed in terms of tangent. We can also express the cotangent function in terms of the sine and cosine.
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