Question -
Answer -
Given:
sin x = 3/5, tan y = 1/2 and π/2 < x< π
We know that, x is in second quadrant and y is inthird quadrant.
In second quadrant, cos x and tan x are negative.
In third quadrant, sec y is negative.
By using the formula,
cos x = – √(1-sin2 x)
tan x = sin x/cos x
Now,
cos x = – √(1-sin2 x)
= – √(1 – (3/5)2)
= – √(1 – 9/25)
= – √((25-9)/25)
= – √(16/25)
= – 4/5
tan x = sin x/cos x
= (3/5)/(-4/5)
= 3/5 × -5/4
= -3/4
We know that sec y = – √(1+tan2 y)
= – √(1 + (1/2)2)
= – √(1 + 1/4)
= – √((4+1)/4)
= – √(5/4)
= – √5/2
Now, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2)
= -6 + 5/2
= (-12+5)/2
= -7/2
∴ 8 tan x – √5 sec y =-7/2