मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Find the values of tan(1050°)

Advertisements
Advertisements

प्रश्न

Find the values of tan(1050°)

बेरीज
Advertisements

उत्तर

tan(1050°) = tan(12 × 90 – 30°)

= – tan30°

= `- 1/sqrt(3)`

shaalaa.com
Trigonometric Functions and Their Properties
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.3 [पृष्ठ १०४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.3 | Q 1. (iv) | पृष्ठ १०४

संबंधित प्रश्‍न

Find the values of `tan ((19pi)/3)`


Prove that cos(π + θ) = − cos θ


Prove that sin(π + θ) = − sin θ.


Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = `pi/2`


If a cos(x + y) = b cos(x − y), show that (a + b) tan x = (a − b) cot y


Prove that sin(n + 1) θ sin(n – 1) θ + cos(n + 1) θ cos(n – 1)θ = cos 2θ, n ∈ Z


Prove that cos 8θ cos 2θ = cos25θ – sin2


If tan x = `"n"/("n" + 1)` and tan y = `1/(2"n" + 1)`, find tan(x + y)


Find the value of cos 2A, A lies in the first quadrant, when cos A = `15/17`


Find the value of cos 2A, A lies in the first quadrant, when sin A = `4/5`


Find the value of cos 2A, A lies in the first quadrant, when tan A  `16/63`


If θ is an acute angle, then find `sin (pi/4 - theta/2)`, when sin θ = `1/25`


Express the following as a sum or difference
sin 4x cos 2x


Express the following as a sum or difference
sin 5θ  sin 4θ


Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x)


Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x


If A + B + C = 180°, prove that `tan  "A"/2  tan  "B"/2 + tan  "B"/2 tan  "C"/2 + tan  "C"/2 tan  "A"/2` = 1


If x + y + z = xyz, then prove that `(2x)/(1 - x^2) + (2y)/(1 - y^2) + (2z)/(1 - z^2) = (2x)/(1 - x^2) (2y)/(1 - y^2) (2z)/(1 - z^2)`


Choose the correct alternative:
`1/(cos 80^circ) - sqrt(3)/(sin 80^circ)` = 


Choose the correct alternative:
If cos 28° + sin 28° = k3, then cos 17° is equal to


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×