Find the value of \[\tan {45^ \circ }\] ?
Answer
550.2k+ views
Hint: We know that tan is a trigonometric function. Also we know that it is the ratio of sin and cos function. So if we know the value of sin and cos for the same angle we can definitely find the value of not only tan function but for all the remaining four functions.
Complete step-by-step answer:
Given that, \[\tan {45^ \circ }\]
We know that,
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Thus we should know the value of sin and cos function for \[{45^ \circ }\] .
We know that, \[\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\] and \[\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]
So taking these value in the ratio we get,
\[\tan {45^ \circ } = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}}\]
Since numerator and denominator are same they are cancelled,
\[\tan {45^ \circ } = 1\]
So, the correct answer is “1”.
Note: This was the simplest method. We can also use a \[{45^ \circ } - {45^ \circ } - {90^ \circ }\] right angle triangle with side opposite to \[{90^ \circ }\] as hypotenuse and remaining two sides as same dimensions say 1. And we know that tan is the ratio of opposite side to adjacent side. That’s the solution!
Complete step-by-step answer:
Given that, \[\tan {45^ \circ }\]
We know that,
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Thus we should know the value of sin and cos function for \[{45^ \circ }\] .
We know that, \[\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\] and \[\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]
So taking these value in the ratio we get,
\[\tan {45^ \circ } = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}}\]
Since numerator and denominator are same they are cancelled,
\[\tan {45^ \circ } = 1\]
So, the correct answer is “1”.
Note: This was the simplest method. We can also use a \[{45^ \circ } - {45^ \circ } - {90^ \circ }\] right angle triangle with side opposite to \[{90^ \circ }\] as hypotenuse and remaining two sides as same dimensions say 1. And we know that tan is the ratio of opposite side to adjacent side. That’s the solution!
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