Find tan 15- and show that tan 15- 15-4

We have, tan 15^∘ = tan (60 45^∘) = tan60^∘ tan45^∘/1 + tan 60 tan 45^∘ [ ∵tan (A B) = tan A tan B/1 + tan A tan B ] =√(3) 1/1+ √(3) = √(3) + 1/√(3) 1 cot ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Equations

Find tan 15∘ ...

Question

Find tan $15^{\circ}$ and show that tan $15^{\circ}$ + cot $15^{\circ}$ = 4

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Solution

We have, $tan 15^{\circ} = tan (60 - 45^{\circ}$ )

= $\frac{tan 60^{\circ} - tan 45^{\circ}}{1 + tan 60 tan 45^{\circ}} [∵ tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}]$

= $\frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$ $cot 15^{\circ} = \frac{1}{tan 15^{\circ}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

and $tan 15^{\circ} + cot 15^{\circ} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

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

= $\frac{2 [(\sqrt{3})^{2} + 1^{2}]}{(\sqrt{3})^{2} - 1} = \frac{2 \times 4}{2} = 4 = RHS$

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Find tan 15∘ and show that tan 15∘ + cot 15∘ = 4

Find tan $15^{\circ}$ and show that tan $15^{\circ}$ + cot $15^{\circ}$ = 4