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

Hint:
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 $\tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} = 1$