How do you find the exact value of \[\tan \left( {\dfrac{\pi }{4}} \right)\]?
Answer
631.2k+ views
Hint: In the given question, we have been asked to calculate the value of a trigonometric ratio. The argument of the trigonometric ratio has also been given. When the denominator of the value of the trigonometric is an irrational number, we shift the irrational part to the numerator by rationalizing the denominator. And then we solve the answer as normal.
Complete step-by-step answer:
We have to calculate the value of \[\tan \left( {\dfrac{\pi }{4}} \right)\].
We know, \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
and, \[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Also, \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
Hence, \[\tan \dfrac{\pi }{4} = \dfrac{{\sin \dfrac{\pi }{4}}}{{\cos \dfrac{\pi }{4}}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}} = 1\]
Thus, \[\tan \dfrac{\pi }{4} = 1\]
Note:We saw that in this question, we did not need to do any effort to rationalize the denominator – the effort is needed only when the denominator is an irrational number.
Complete step-by-step answer:
We have to calculate the value of \[\tan \left( {\dfrac{\pi }{4}} \right)\].
We know, \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
and, \[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Also, \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
Hence, \[\tan \dfrac{\pi }{4} = \dfrac{{\sin \dfrac{\pi }{4}}}{{\cos \dfrac{\pi }{4}}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}} = 1\]
Thus, \[\tan \dfrac{\pi }{4} = 1\]
Note:We saw that in this question, we did not need to do any effort to rationalize the denominator – the effort is needed only when the denominator is an irrational number.
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