Question

If 4sin2 theta+4cosec2 theta=17 , then the value of tan theta is -​

Answer 1

Answer:

tanθ = 7.873 , 1.873

Step-by-step explanation:

Given ,

4sin2θ+ 4cosec2θ = 17

To Find :-

Value of 'tanθ'

How To Do :-

We need to convert cosec2θ  in terms of sin2θ  and we need to simplify it after simplifying it we can observe that it forms a quadratic equation with variable 'sin2θ ' we need to solve it by using splitting the middle term and we need to obtain the value of sin2θ  and we need to equate that to formula sin2θ  in terms of tanθ  and we need to simplify it to obtain the value of tanθ .

Formula Required :-

1) cosec2θ  = 1/sin2θ

2) sinθ  ∈ [1 , - 1]

3) sin2θ  = 2tanθ /(1 + tan²θ )

4) tanθ  ∈ (-∝ , ∝ )

Solution :-

4sin2θ + 4cosec2θ = 17

Taking '4' as common :-

4(sin2θ + cosec2θ) = 17

[ ∴ cosec2θ  = 1/sin2θ  ]

4(sin2θ + 1/sin2θ) = 17

Taking L.C.M :-

4[(sin2θ)(sin2θ)+1/sin2θ] = 17

4[(sin²2θ + 1)/sin2θ] = 17

(4sin²2θ + 4)/sin2θ = 17

4sin²2θ + 4 = 17(sin2θ)

4sin²2θ + 4 = 17sin2θ

4sin²2θ - 17sin2θ + 4 = 0

Solving by using splitting the term method :-

4sin²2θ - 16sin2θ - sin2θ + 4 = 0

4sin2θ(sin2θ - 4) - 1(sin2θ - 4) = 0

Taking 'sin2θ - 4' as common :-

(sin2θ - 4)(4sin2θ - 1) = 0

Equating both terms to '0' :-

(sin2θ - 4) = 0 , (4sin2θ - 1) = 0

Equating 'sin2θ - 4' to '0' :-

sin2θ - 4 = 0

sin2θ = 4

sin2θ ≠ 4 because the 'sin' ratio lies in between '-1' and '1'

[ ∴ sinθ  ∈ [1 , - 1]

Equating '4sin2θ - 1' to '0' :-

4sin2θ - 1 = 0

4sin2θ = 1

sin2θ = 1/4

[ ∴ sin2θ  = 2tanθ /(1 + tan²θ ) ]

2tanθ/ (1 + tan²θ) = 1/4

Cross multiplication :-

4(2tanθ) = 1(1 + tan²θ)

8tanθ = 1 + tan²θ

tan²θ - 8tanθ + 1 = 0

Solving by using quadratic formula :-

$tan\theta=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(1)(1)}}{2(1)}$

$=\dfrac{8\pm\sqrt{64-4}}{2}$

$=\dfrac{8\pm\sqrt{60}}{2}$

$=\dfrac{8\pm\sqrt{15\times 4}}{2}$

$=\dfrac{8\pm2\sqrt{15}}{2}$

$=\dfrac{2(4\pm\sqrt{15})}{2}$

= 4 ± √15

= 4 ± 3.873

= 4 + 3.873 , 4 - 3.873

= 7.873 , 1.873

tanθ = 7.873 , 1.873

Because both the numbers satisfies the range of 'tan' ratio.

∴ tan ∈ (-∝ , ∝)