Answer :
For tangent 105 degrees:
Using the tangent sum identity:
tan(105°) = (tan(45°) + tan(60°)) / (1 - tan(45°) * tan(60°))
tan(45°) = 1
tan(60°) = √3
Substituting these values:
tan(105°) = (1 + √3) / (1 - √3)
Rationalizing the denominator:
tan(105°) = (1 + √3)(1 + √3) / (1 - √3)(1 + √3)
tan(105°) = (1 + 2√3 + 3) / (1 - 3)
tan(105°) = (4 + 2√3) / (-2)
tan(105°) = -(2 + √3)
For tangent (5π)/12:
Using the tangent difference identity:
tan(5π/12) = (tan(π/3) - tan(π/4)) / (1 + tan(π/3) * tan(π/4))
tan(π/3) = √3
tan(π/4) = 1
Substituting these values:
tan(5π/12) = (√3 - 1) / (1 + √3 * 1)
tan(5π/12) = (√3 - 1) / (1 + √3)
Rationalizing the denominator:
tan(5π/12) = (√3 - 1)(1 - √3) / (1 + √3)(1 - √3)
tan(5π/12) = (√3 - √3 - 1 + 1) / (1 - 3)
tan(5π/12) = (-√3) / (-2)
tan(5π/12) = √3/2
Answer:
5) [tex]\sf \tan 105^\circ = -2 - \sqrt{3} [/tex]
6) [tex]\sf \tan \dfrac{5\pi}{12} = 2 + \sqrt{3} [/tex]
Step-by-step explanation:
To find the exact values of [tex]\sf \tan 105^\circ [/tex] and [tex]\sf \tan \dfrac{5\pi}{12} [/tex], we can use sum and difference identities along with known trigonometric values. Let's solve each one step by step.
5) [tex]\sf \tan 105^\circ [/tex]
First, we'll use the identity [tex]\sf \tan(180^\circ - \theta) = -\tan(\theta) [/tex] to rewrite [tex]\sf \tan 105^\circ [/tex]:
[tex]\sf \tan 105^\circ = \tan(180^\circ - 75^\circ) = -\tan 75^\circ [/tex]
Next, we'll use the identity [tex]\sf \tan(a + b) = \dfrac{\tan a + \tan b}{1 - \tan a \tan b} [/tex] with [tex]\sf a = 45^\circ [/tex] and [tex]\sf b = 30^\circ [/tex]
(since [tex]\sf 75^\circ = 45^\circ + 30^\circ [/tex]):
[tex]\sf \tan 75^\circ = \tan(45^\circ + 30^\circ) = \dfrac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} [/tex]
Using known values:
[tex]\sf \tan 45^\circ = 1 [/tex]
[tex]\sf \tan 30^\circ = \dfrac{1}{\sqrt{3}} [/tex]
Substitute these values:
[tex]\sf \tan 75^\circ = \dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - 1 \cdot \dfrac{1}{\sqrt{3}}} [/tex]
Simplify further:
[tex]\sf \tan 75^\circ = \dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{\sqrt{3}}} [/tex]
To rationalize the denominator, multiply numerator and denominator by [tex]\sf \sqrt{3} [/tex]:
[tex]\sf \tan 75^\circ = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} [/tex]
To remove the square root from the denominator, multiply numerator and denominator by the conjugate:
[tex]\sf \tan 75^\circ = \dfrac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} [/tex]
[tex]\sf \tan 75^\circ = \dfrac{3 + 2\sqrt{3} + 1}{3 - 1} [/tex]
[tex]\sf \tan 75^\circ = \dfrac{4 + 2\sqrt{3}}{2} [/tex]
[tex]\sf \tan 75^\circ = 2 + \sqrt{3} [/tex]
Therefore,
[tex]\sf \tan 105^\circ = -\tan 75^\circ = - (2 + \sqrt{3}) = -2 - \sqrt{3} [/tex]
6) [tex]\sf \tan \dfrac{5\pi}{12} [/tex]
For [tex]\sf \tan \dfrac{5\pi}{12} [/tex], we'll use the sum and difference identities along with the known values of trigonometric ratios:
[tex]\sf \tan \dfrac{5\pi}{12} = \tan \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} \right) [/tex]
Using the identity [tex]\sf \tan(a + b) = \dfrac{\tan a + \tan b}{1 - \tan a \tan b} [/tex] with [tex]\sf a = \dfrac{\pi}{4} [/tex] and [tex]\sf b = \dfrac{\pi}{6} [/tex]:
[tex]\sf \tan \dfrac{5\pi}{12} = \dfrac{\tan \dfrac{\pi}{4} + \tan \dfrac{\pi}{6}}{1 - \tan \dfrac{\pi}{4} \tan \dfrac{\pi}{6}} [/tex]
Using known values:
[tex]\sf \tan \dfrac{\pi}{4} = 1 [/tex]
[tex]\sf \tan \dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}} [/tex]
Substitute these values:
[tex]\sf \tan \dfrac{5\pi}{12} = \dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - 1 \cdot \dfrac{1}{\sqrt{3}}} [/tex]
Rationalize the denominator:
[tex]\sf \tan \dfrac{5\pi}{12} = \dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{\sqrt{3}}} [/tex]
Multiply numerator and denominator by [tex]\sf \sqrt{3} [/tex] to rationalize:
[tex]\sf \tan \dfrac{5\pi}{12} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} [/tex]
Multiply numerator and denominator by the conjugate:
[tex]\sf \tan \dfrac{5\pi}{12} = \dfrac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} [/tex]
[tex]\sf \tan \dfrac{5\pi}{12} = \dfrac{3 + 2\sqrt{3} + 1}{3 - 1} [/tex]
[tex]\sf \tan \dfrac{5\pi}{12} = \dfrac{4 + 2\sqrt{3}}{2} [/tex]
[tex]\sf \tan \dfrac{5\pi}{12} = 2 + \sqrt{3} [/tex]
Therefore,
[tex]\sf \tan \dfrac{5\pi}{12} = 2 + \sqrt{3} [/tex]
In conclusion:
5) [tex]\sf \tan 105^\circ = -2 - \sqrt{3} [/tex]
6) [tex]\sf \tan \dfrac{5\pi}{12} = 2 + \sqrt{3} [/tex]
