The tangent of 3 radians is approximately -0.427. This value can be calculated using a scientific calculator or a trigonometric table. In degrees, the tangent of 3 radians is equivalent to the tangent of about 171.89 degrees.
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cot(15)=1/tan(15) Let us find tan(15) tan(15)=tan(45-30) tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b)) tan(45-30)= (tan(45)-tan(30))/(1+tan(45)tan(30)) substitute tan(45)=1 and tan(30)=1/√3 into the equation. tan(45-30) = (1- 1/√3) / (1+1/√3) =(√3-1)/(√3+1) The exact value of cot(15) is the reciprocal of the above which is: (√3+1) /(√3-1)
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If the angles are measured in degrees or gradians, then: tan 3 > tan 2 > tan 1 If the angles are measured in radians, then: tan 1 > tan 3 > tan 2.
tan(pi/3)= sqrt(3)
To find the exact value of tan 105°. First, of all, we note that sin 105° = cos 15°; and cos 105° = -sin 15°. Thus, tan 105° = -cot 15° = -1 / tan 15°. Using the formula tan(α - β) = (tan α - tan β) / (1 + tan α tan β); and using, also, the familiar values tan 45° = 1, and tan 30° = ½ / (½√3) = 1/√3 = ⅓√3; we have, tan 15° = (1 - ⅓√3) / (1 + ⅓√3); whence, cot 15° = (1 + ⅓√3) / (1 - ⅓√3) = (√3 + 1) / (√3 - 1) {multiplying through by √3} = (√3 + 1)2 / (√3 + 1)(√3 - 1) = (3 + 2√3 + 1) / (3 - 1) = (4 + 2√3) / 2 = 2 + √3. Therefore, tan 105° = -cot 15° = -2 - √3, which is the result we sought. We are asked the exact value of tan 105°, which we gave above. We can test the above result to 9 decimal places, say, by means of a calculator: -2 - √3 = -3.732050808; and tan 105° = -3.732050808; thus indicating that we have probably got the right result.
tan θ = √3. tan-1(√3) = π/3 radians or 4π/3 radians, in degrees it would be 60° or 240° tan θ = sin θ / cos θ So, tan (π/3) = sin (π/3) / cos (π/3) = tan (4π/3) = sin (4π/3) / cos (4π/3) cos (π/3) = 1/2 cos (4π/3) = -1/2 Therefore cos θ = 1/2 or -1/2 when tan θ = √3
tan(60°) = sqrt(3)
Tan[sqrt(3) radians] = tan(1.7321) = -6.1475
tan(30)=.5773502692
The value of ( \tan 15^\circ ) can be calculated using the tangent subtraction formula: [ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} ] Substituting the known values ( \tan 45^\circ = 1 ) and ( \tan 30^\circ = \frac{1}{\sqrt{3}} ), we find: [ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} ] Thus, ( \tan 15^\circ = 2 - \sqrt{3} ) in its simplest fractional form.
tan(pi/3) = tan (60 degrees) = 1.732 which is square root of 3
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