Is the '1' one degree , or one radian or, Tan(angle) = 1.
You need to clarify.
However,
Tan(1 degree) = 0.017455...
Tan( 1 radian) = 1.55740....
Tan(angle) = 1
Angle = Tan^(-1) 1 = 45 degrees. = pi/4 radians.
Any number can be a tan. So -sqrt(17), 19.56, 45678942 are all examples of tan. Cosine can have any value in the range [-1, 1].
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.
1/sqrt(3)
To find the value of (\tan(15^\circ) \tan(195^\circ)), we can use the identity (\tan(195^\circ) = \tan(15^\circ + 180^\circ) = \tan(15^\circ)). Thus, (\tan(195^\circ) = \tan(15^\circ)). Consequently, (\tan(15^\circ) \tan(195^\circ) = \tan(15^\circ) \tan(15^\circ) = \tan^2(15^\circ)). The exact value of (\tan^2(15^\circ)) can be computed, but it is important to note that it will yield a positive value.
1 because tan(5 pi / 4) = 1
tan (pi) / 1 is zero. tan (pi / 1) is zero.
tan(135) = -tan(180-135) = -tan(45) = -1
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)
The value of tan A is not clear from the question.However, sin A = sqrt[tan^2 A /(tan^2 A + 1)]
tan(135 degrees) = negative 1.
1
Any number can be a tan. So -sqrt(17), 19.56, 45678942 are all examples of tan. Cosine can have any value in the range [-1, 1].
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.
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.
The value of tan and sin is positive so you must search quadrant that tan and sin value is positive. The only quadrant fill that qualification is Quadrant 1.
1/sqrt(3)
tan u/2 = sin u/1+cos u