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

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\\} = 1if $n = 0$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\\} = 1if $n = 2$ 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\\} = 1if $n = 3$ 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 ${0^ \circ } - {90^ \circ }$&${180^ \circ } - {270^ \circ }$.