Ex 3.2, 4 - Find values of other five functions if sec x = 13/5 - Ex 3.2

part 2 - Ex 3.2, 4 - Ex 3.2 - Serial order wise - Chapter 3 Class 11 Trigonometric Functions
part 3 - Ex 3.2, 4 - Ex 3.2 - Serial order wise - Chapter 3 Class 11 Trigonometric Functions

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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)^2– 1 tan2x = 169/25 – 1 tan2x = (169 βˆ’ 25)/25 tan2x = 144/25 tan x = Β± √(144/25) tan x = Β± 𝟏𝟐/πŸ“ Since x is in lVth Quadrant tan x is negative in lVth Quadrant ∴ tan x = (βˆ’πŸπŸ)/πŸ“ cot x = 1/tπ‘Žπ‘›β‘π‘₯ = 1/(" " (βˆ’12)/5) = (βˆ’πŸ“)/𝟏𝟐 tan x = sin⁑π‘₯/cos⁑π‘₯ sin x = (tan x) Γ— (cos x ) = (βˆ’12)/5 Γ— 5/13 = (βˆ’πŸπŸ)/πŸπŸ‘ cos x = 1/s𝑒𝑐⁑π‘₯ = 1/(" " 13/5) = πŸ“/πŸπŸ‘ cosec x = 1/sin⁑π‘₯ = (βˆ’πŸπŸ‘)/𝟏𝟐 Ex 3.2, 5 Find the values of other five trigonometric functions if tan⁑π‘₯ = βˆ’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 + ((βˆ’5)/12)^2 = sec2x 1 + 25/144 = sec2x (144 + 25)/144 = sec2x 169/144 = sec2x sec2x = 169/144 sec2x = πŸπŸ”πŸ—/πŸπŸ’πŸ’ sec x = Β± √(169/144) sec x = Β± πŸπŸ‘/𝟏𝟐 As x is in llnd Quadrant, cos x is negative in IInd quadrant

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