Question -
Answer -
Given:
sin x = 3/5, tan y = 1/2 and┬а╧А/2 < x< ╧А
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 = тАУ тИЪ(1-sin2┬аx)
tan x = sin x/cos x
Now,
cos x = тАУ тИЪ(1-sin2┬аx)
= тАУ тИЪ(1 тАУ (3/5)2)
= тАУ тИЪ(1 тАУ 9/25)
= тАУ тИЪ((25-9)/25)
= тАУ тИЪ(16/25)
= тАУ 4/5
tan x = sin x/cos x
= (3/5)/(-4/5)
= 3/5 ├Ч -5/4
= -3/4
We know that sec y = тАУ тИЪ(1+tan2┬аy)
= тАУ тИЪ(1 + (1/2)2)
= тАУ тИЪ(1 + 1/4)
= тАУ тИЪ((4+1)/4)
= тАУ тИЪ(5/4)
= тАУ тИЪ5/2
Now, 8 tan x тАУ тИЪ5 sec y = 8(-3/4) тАУ тИЪ5(-тИЪ5/2)
= -6 + 5/2
= (-12+5)/2
= -7/2
тИ┤ 8 tan x тАУ тИЪ5 sec y =-7/2