Question

If sinx +cosx = root 7 by 2 where x E [0,45] then what will be the value of tan x by 2?

Answer 1

Sinx + cosx = √7/2 where x∈[0, 45°]

sinx + cosx = √7/2
We know, sin2Ф = 2tanФ/(1 + tan²Ф)
cos2Ф = (1 - tan²Ф)/(1 + tan²Ф)
Now, sinx + cosx = √7/2
⇒2tanx/2/(1 + tan²x/2) + (1 - tan²x/2)/(1 + tan²x/2) =√7/2
⇒(1 - tan²x/2 + 2tanx/2)/(1 + tan²x/2) = √7/2
⇒2 - 2tan²x/2 + 4tanx/2 = √7 + √7tan²x/2
⇒(2 + √7) tan²x/2 - 4tanx/2 + (-2 + √7) = 0
tanx/2 = {4 ± √(16 - 12)}/2(2 + √7)
tanx/2 = (2 ± 1)/(2 + √7)
Because x ∈ [0, 45°]
so, x/2 ∈ [0, 22.5°]
but tan22.5° = √(1 - 1/√2)/(1 + 1/√2) =√(√2 - 1)/(√2 + 1) = √2 -1 = 0.414
so, 0 ≤ tanx/2 ≤ 0.414

∴tanx/2 ≠ 3/(2 + √7)

Hence, tanx/2 = 1/(2 + √7)

Answer 2

Answer:

root 7-2/3

Step-by-step explanation:

Sinx + cosx = √7/2 where x∈[0, 45°]

sinx + cosx = √7/2

We know, sin2Ф = 2tanФ/(1 + tan²Ф)

cos2Ф = (1 - tan²Ф)/(1 + tan²Ф)

Now, sinx + cosx = √7/2

⇒2tanx/2/(1 + tan²x/2) + (1 - tan²x/2)/(1 + tan²x/2) =√7/2

⇒(1 - tan²x/2 + 2tanx/2)/(1 + tan²x/2) = √7/2

⇒2 - 2tan²x/2 + 4tanx/2 = √7 + √7tan²x/2

⇒(2 + √7) tan²x/2 - 4tanx/2 + (-2 + √7) = 0

tanx/2 = {4 ± √(16 - 12)}/2(2 + √7)

tanx/2 = (2 ± 1)/(2 + √7)

Because x ∈ [0, 45°]

so, x/2 ∈ [0, 22.5°]

but tan22.5° = √(1 - 1/√2)/(1 + 1/√2) =√(√2 - 1)/(√2 + 1) = √2 -1 = 0.414

so, 0 ≤ tanx/2 ≤ 0.414

∴tanx/2 ≠ 3/(2 + √7)

Hence, tanx/2 = 1/(2 + √7