1.

If sin x = 3/5, tan y = 1/2 and π/2 < x < π < y < 3π/2 find the value of 8 tan x - √5 sec y.

Answer»

Given as

sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 

As we know that, x is in second quadrant and y is in third quadrant.

Since, in second quadrant, cos x and tan x are negative.

Since, in third quadrant, sec y is negative.

On using the formula,

cos x = – √(1 - sin2 x)

tan x = sin x/cos x

Then,

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

As we know that sec y = – √(1 + tan2 y)

= – √(1 + (1/2)2)

= – √(1 + 1/4)

= – √((4 + 1)/4)

= – √(5/4)

= – √5/2

Then, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2)

= -6 + 5/2

= (-12 + 5)/2

= -7/2

Thus, 8 tan x – √5 sec y = -7/2



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