Ex 3.2

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Ex 3.2, 4 Find the values of other five trigonometric functions if sec x = 13/5 , π₯ lies in fourth quadrant. Since x lies in the lVth Quadrant Where cos will be positive But sin and tan will be negative We know that 1 + tan2x = sec2x 1 + tan2x = (13/5)^2 tan2x = (13/5)^2β 1 tan2x = 169/25 β 1 tan2x = (169 β 25)/25 tan2x = 144/25 tan x = Β± β(144/25) tan x = Β± ππ/π Since x is in lVth Quadrant tan x is negative in lVth Quadrant β΄ tan x = (βππ)/π cot x = 1/tππβ‘π₯ = 1/(" " (β12)/5) = (βπ)/ππ tan x = sinβ‘π₯/cosβ‘π₯ sin x = (tan x) Γ (cos x ) = (β12)/5 Γ 5/13 = (βππ)/ππ cos x = 1/sππβ‘π₯ = 1/(" " 13/5) = π/ππ cosec x = 1/sinβ‘π₯ = (βππ)/ππ Ex 3.2, 5 Find the values of other five trigonometric functions if tanβ‘π₯ = β5/12 , π₯ lies in second quadrant. Since x lies in llnd Quadrant So, sin x will be positive But tan x and cos x will be negative We know that 1 + tan2x = sec2x 1 + ((β5)/12)^2 = sec2x 1 + 25/144 = sec2x (144 + 25)/144 = sec2x 169/144 = sec2x sec2x = 169/144 sec2x = πππ/πππ sec x = Β± β(169/144) sec x = Β± ππ/ππ As x is in llnd Quadrant, cos x is negative in IInd quadrant