Question

12
If 8 sino = 4 + coso, find the value of tano.​

Answer 1

Step-by-step explanation:

Given that

8sinθ=4+cosθ

8sinθcosθ=4+cosθcosθ

8tanθ=4secθ+1

4secθ=8tanθ−1

(4secθ)2=(8tanθ−1)2

16sec2θ=64tan2θ−16tanθ+1

16(1+tan2θ)=64tan2θ−16tanθ+1

16+16tan2θ=64tan2θ−16tanθ+1

48tan2θ−16tanθ−15=0

solving above quadratic equation for tanθ as follows

tanθ=−(−16)±(−16)2−4(48)(−15)√2(48)

tanθ=16±5696

tanθ=34 or −512

Both the above values of tanθ are acceptable as they duly satisfy the (original) given equation: 8sinθ=4+cosθ

Answer 2

Alternatively:—

Let theta=x

8sinx=4+cosx

Divide both side by cosx

8tanx=4secx+1

8 tanx -1= 4 secx , squaring both sides

64 tan^2x -16 tanx +1 =16sec^2x

64 tan^2x -16tanx +1= 16(1+tan^2x)

48tan^2x -16 tanx -15 =0

48 tan^2x-36tanx+20tanx-15=0

12tanx(4tanx-3)+5(4tanx-3)=0

(4tanx-3)(12tanx+5)=0

Either 4tanx-3=0 => tanx=3/4

Or 12tanx+5=0 => tanx = -5/12.

tan x= 3/4 , - 5/12

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