Question

Find the values of other five trigonometric functions. tanx = √3, where x lies in third quadrant.

Answer 1

Answer:

(i) Since, in third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative. On using the formulas, tan x = 1/cot x = 1/(12/5) = 5/12 cosec x = –√(1 + cot2 x) = –√(1 + (12/5)2) = –√(25 + 144)/25 = –√(169/25) = -13/5 sin x = 1/cosec x = 1/(-13/5) = -5/13 cos x = – √(1 – sin2 x) = – √(1 – (-5/13)2) = – √(169 - 25)/169 = – √(144/169) = -12/13 sec x = 1/cos x = 1/(-12/13) = -13/12 ∴ sin x = -5/13, cos x = -12/13, tan x = 5/12, cosec x = -13/5, sec x = -13/12 (ii) Since, in second quadrant, sin x and cosec x are positive. tan x, cot x, cos x, sec x are negative. On using the formulas, sin x = √(1 – cos2 x) = √(1 – (-1/2)2) = √(4 - 1)/4 = √(3/4) = √3/2 tan x = sin x/cos x = (√3/2)/(-1/2) = -√3 cot x = 1/tan x = 1/-√3 = -1/√3 cosec x = 1/sin x = 1/(√3/2) = 2/√3 sec x = 1/cos x = 1/(-1/2) = -2 ∴ sin x = √3/2, tan x = -√3, cosec x = 2/√3, cot x = -1/√3 sec x = -2 (iii) Since, in third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative. On using the formulas, sin x = √(1 – cos2 x) = – √(1 - (-4/5)2) = – √(25 - 16)/25 = – √(9/25) = – 3/5 cos x = 1/sec x = 1/(-5/4) = -4/5 cot x = 1/tan x = 1/(3/4) = 4/3 cosec x = 1/sin x = 1/(-3/5) = -5/3 sec x = -√(1 + tan2 x) = – √(1 + (3/4)2) = – √(16 + 9)/16 = – √ (25/16) = -5/4 ∴ sin x = -3/5, cos x = -4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3 (iv) Since, in first quadrant, all trigonometric ratios are positive. Therefore, by using the formulas, tan x = sin x/cos x = (3/5)/(4/5) = 3/4 cosec x = 1/sin x = 1/(3/5) = 5/3 cos x = √(1 - sin2 x) = √(1 – (-3/5)2) = √(25 - 9)/25 = √(16/25) = 4/5 sec x = 1/cos x = 1/(4/5) = 5/4 cot x = 1/tan x = 1/(3/4) = 4/3 Thus, cos x = 4/5, tan x = 3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3Read more on Sarthaks.com - https://www.sarthaks.com/652756/find-the-values-the-other-five-trigonometric-functions-each-the-following-cot-quadrant-iii