We have tan 13 π/12 = tan ( π + π/12 ) = tan π/12 [ ∵ tan (π + θ) = tan θ ] = tan ( π/3 π/4 ) = tan π/3 tan π/4/1 + tan π/3 tan π/4 = √(3) 1/1 + √(3) = ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Equations

Evaluate tan ...

Question

Evaluate tan $\frac{13 π}{12}$

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Solution

We have

tan $\frac{13 π}{12}$ = tan $(π + \frac{π}{12})$

= $t a n \frac{π}{12} [∵ t a n (π + θ) = t a n θ]$

= $t a n (\frac{π}{3} - \frac{π}{4})$

= $\frac{t a n \frac{π}{3} - t a n \frac{π}{4}}{1 + t a n \frac{π}{3} t a n \frac{π}{4}} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}$

= $\frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)} \times \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^{2}}{(3 - 1)}$

= $\frac{(3 + 1 - 2 \sqrt{3})}{2} = \frac{(4 - 2 \sqrt{3})}{2} = (2 - \sqrt{3})$

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Evaluate tan 13π12

We have tan 13π12 = tan (π+π12) = tanπ12[∵tan(π+θ)=tan θ] = tan(π3−π4) = tanπ3−tanπ41+ tan π3 tan π4=√3−11+√3 = (√3−1)(√3+1)×(√3−1)(√3−1)=(√3−1)2(3−1) = (3+1−2√3)2=(4−2√3)2=(2−√3)

Evaluate tan $\frac{13 π}{12}$