Question

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

Answer 1

Hint: n is an integer which refers to the numbers which include whole numbers, positive numbers and negative numbers. An integer number cannot have a fraction or decimal. Integer numbers can be \[n = .........., - 2, - 1,0,1,2,3.........\]
In this question check the value of the given function weather it is equal to 1 by substituting the n with integer numbers.


Complete step by step solution:
Given \[n\] is an integer number, so check for the value of the trigonometric function weather it is equal to 1 when integer values are substituted
Case 1: When \[n = 0\]by putting this integer value we get
\[
  \tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {\dfrac{{0 \times \pi }}{2} + {{\left( { - 1} \right)}^0}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {0 + \dfrac{\pi }{4}} \right\} \\
  \tan \left( {\dfrac{\pi }{4}} \right) = 1 \\
 \]
Hence we can say \[\tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} = 1\]if \[n = 0\]since \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
Case 2: When\[n = 1\]by putting this integer value we get
\[
  \tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {\dfrac{{1 \times \pi }}{2} + {{\left( { - 1} \right)}^1}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right\} \\
  \tan \left( {\dfrac{\pi }{4}} \right) = 1 \\
 \]
Hence we can say \[\tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} = 1\] if \[n = 1\] since \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
Case 3: When \[n = 2\] by putting this integer value we get
\[
  \tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {\dfrac{{2 \times \pi }}{2} + {{\left( { - 1} \right)}^2}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {\pi + \dfrac{\pi }{4}} \right\} \\
  \tan \left( {\dfrac{\pi }{4}} \right) = 1 \\
 \]
Hence, we can say \[\tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} = 1\]if \[n = 2\] since \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
Case 4: When\[n = 3\] by putting this integer value we get
\[
  \tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {\dfrac{{3 \times \pi }}{2} + {{\left( { - 1} \right)}^3}\dfrac{\pi }{4}} \right\} \\
  \tan \left\{ {\dfrac{{3\pi }}{2} - \dfrac{\pi }{4}} \right\} \\
  \tan \left( {\dfrac{{5\pi }}{4}} \right) = 1 \\
 \]
Hence, we can say \[\tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} = 1\]if \[n = 3\] since \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
So the trigonometric function \[\tan \left\{ {\dfrac{{n\pi }}{2} + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right\} = 1\] when the values of integers are \[n = 0,1,2,3.........\infty \]


Note: Trigonometric functions have different values when they lie in the four quadrants.
The trigonometric values of a tan function are positive when they lie in the first and third quadrants, which are ranging from \[{0^ \circ } - {90^ \circ }\]&\[{180^ \circ } - {270^ \circ }\].