Question

If n is an integer, prove that
\tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\} = 1

Answer 1

It is given in the question If n is an integer. Then, we will prove that
\tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\} = 1
By putting values of n = 1, 2, 3 and check for every individual value whether the value satisfies the equation or not.

Complete step by step solution:
It is given in the question If n is an integer. Then, we have to prove that
\tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\} = 1
Now, put n = 1 in the above equation
= \tan \left\\{ {\dfrac{\pi }{2} + \left( { - 1} \right)\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right\\}
$= \tan \dfrac{\pi }{4}$
Since, we know that $\tan \dfrac{\pi }{4}$ is equal to 1.
$\therefore$ =1
Thus, for $n = 1$ the given equation is satisfied.
Now, taking $n = 2$
= \tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {\dfrac{{2\pi }}{2} + {{\left( { - 1} \right)}^2}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {\pi + \dfrac{\pi }{4}} \right\\}
Since, we know that $\pi + \dfrac{\pi }{4}$ lies in third quadrant and in third quadrant tan function is positive.
$= \tan \dfrac{\pi }{4}$
=1
Thus, for $n = 2$ the given equation is satisfied.
Now, putting $n = 3$
= \tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {\dfrac{{3\pi }}{2} + {{\left( { - 1} \right)}^3}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {\dfrac{{3\pi }}{2} - \dfrac{\pi }{4}} \right\\}
Since, we know that $\dfrac{{3\pi }}{2} - \dfrac{\pi }{4}$ lies in third quadrant and in third quadrant tan function is positive.
$= \tan \dfrac{\pi }{4}$
=1
Thus, for $n = 3$ the given equation is satisfied.

Hence proved \tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\} = 1

Note:
The above question can be solved with another method:
It is given in the question If n is an integer. Then, we have to prove that
\tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\} = 1
Now, put 2k in the above equation.
= \tan \left\\{ {\dfrac{{2k\pi }}{2} + {{\left( { - 1} \right)}^{2k}}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {k\pi + {{\left( { - 1} \right)}^{2k}}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {k\pi + \dfrac{\pi }{4}} \right\\} for every $k \in Z$
For every $k \in Z$ , \tan \left\\{ {k\pi + \dfrac{\pi }{4}} \right\\} is equal to 1.
=1
Thus, for n = 2k the given equation is satisfied.
Now, taking $n = 2k + 1$ ,
= \tan \left\\{ {\dfrac{{\left( {2k + 1} \right)\pi }}{2} + {{\left( { - 1} \right)}^{2k + 1}}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {\dfrac{{\left( {2k\pi + \pi } \right)}}{2} + {{\left( { - 1} \right)}^{2k + 1}}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {\dfrac{{2k\pi }}{2} + \dfrac{\pi }{2} + {{\left( { - 1} \right)}^{2k}}{{\left( { - 1} \right)}^1}\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {k\pi + \dfrac{\pi }{2} + \left( {1 \times - 1} \right)\dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {k\pi + \dfrac{\pi }{2} - \dfrac{\pi }{4}} \right\\}
= \tan \left\\{ {k\pi + \dfrac{\pi }{4}} \right\\}
=1
Thus, for $n = 2k + 1$ the given equation is satisfied.
Hence proved \tan \left\\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\\} = 1