Question

Tan(45+theta)+tan(45-theta)/Tan(45+theta)-tan(45-theta)

Answer 1

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Given,

• $\dfrac{tan(45+\theta)+tan(45-\theta)}{tan(45+\theta)-tan(45-\theta)}$

$\large{\blue{\fbox{\orange{\underline{\red{\star\;Note}}}}}}$

• $tan(a+b)=\dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$

• $tan(a-b)=\dfrac{tan(a)-tan(b)}{1+tan(a)tan(b)}$

• $1+tan^2(\theta)=sec^2(\theta)$

Substituting in the formula , we get,

• $\dfrac{tan(45+\theta)+tan(45-\theta)}{tan(45+\theta)-tan(45-\theta)}$

• $\dfrac{\frac{tan(45)+tan(\theta)}{1-tan(45)tan(\theta)} + \frac{tan(45)-tan(\theta)}{1+tan(45)tan(\theta)}}{\frac{tan(45)+tan(\theta)}{1-tan(45)tan(\theta)}-\frac{tan(45)-tan(\theta)}{1+tan(45)tan(\theta)}}$

• $\dfrac{\frac{1+tan(\theta)}{1-tan(\theta)} + \frac{1-tan(\theta)}{1+tan(\theta)}}{\frac{1+tan(\theta)}{1-tan(\theta)}-\frac{1-tan(\theta)}{1+tan(\theta)}}$

• $\dfrac{{(1+tan(\theta))}^2 + {(1-tan(\theta))}^2}{{(1+tan(\theta))}^2 - {(1-tan(\theta))}^2}$

• $\dfrac{1+tan^2(\theta)+2tan(\theta)+1+tan^2(\theta)-2tan(\theta)}{1+tan^2(\theta)+2tan(\theta)-(1+tan^2(\theta)-2tan(\theta))}$

• $\dfrac{2+2tan^2(\theta)}{4tan(\theta)}$

• $\dfrac{2sec^2(\theta)}{4tan(\theta)}$

• $\dfrac{sec^2(\theta)}{2tan(\theta)}$