If f(θ)=tan⁡θ and f(a)=2, find the exact value of: a) f(−a) b) f(a)+f(a+π)+f(a+2π)

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If f(θ)=tan⁡θ and f(a)=2, find the exact value of:  a) f(−a)  b) f(a)+f(a+π)+f(a+2π)

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1)&amp;nbsp;f(θ)=tan⁡θ&amp;nbsp;2)&amp;nbsp;f(a)=2&amp;nbsp;a) To find&amp;nbsp;f(−a):&amp;nbsp;Substitute&amp;nbsp;(−θ)&amp;nbsp;in place of&amp;nbsp;θ&amp;nbsp;in (1):&amp;nbsp;f(−θ)=tan⁡(−θ)&amp;nbsp;From the tangent trigonometric function&#039;s characteristic:&amp;nbsp;tan⁡(−θ)=−tan⁡θ&amp;nbsp;Thus,&amp;nbsp;3)&amp;nbsp;f(−θ)=−tan⁡θ=−f(θ)&amp;nbsp;Equation (3) is the result of the odd function condition.So,&amp;nbsp;f(−a)=−f(a)=−2&amp;nbsp;&amp;nbsp;b) According to other property of tangent trigonometric function:&amp;nbsp;tan⁡θ=tan⁡(θ+π)&amp;nbsp;i.e. Period of&amp;nbsp;tan⁡&amp;nbsp;θ&amp;nbsp;is&amp;nbsp;π&amp;nbsp;which implies:&amp;nbsp;tan⁡θ=tan⁡(θ+π)=tan⁡(θ+2π)=tan⁡(θ+3π)=⋯=tan⁡(θ+nπ)&amp;nbsp;Where n is an integer.&amp;nbsp;Hence from (1):&amp;nbsp;f(θ)=f(θ+π)=f(θ+2π)=f(θ+3π)=⋯=f(θ+nπ)&amp;nbsp;Substitute&amp;nbsp;θ=a:&amp;nbsp;f(a)=f(a+π)=f(a+2π)=f(a+3π)=⋯=f(a+nπ)&amp;nbsp;Using (2):&amp;nbsp;f(a)=f(a+π)=f(a+2π)=f(a+3π)=⋯=f(a+nπ)=2&amp;nbsp;Then,&amp;nbsp;f(a)+f(a+π)+f(a+2π)=3f(a)&amp;nbsp;3⋅2&amp;nbsp;=6

If f(\theta)=\tan\theta and f(a)=2, find the exact value of: a) f(-a)ZS

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