solve this trigonometric problem If theta=330°, then using the formula tan theta={2tan (theta/(2)}/}1-tan^(2)(theta)/(2)} find the value of {tan(theta)/(2)}.
Answers
Answer:-
Given:-
θ = 330°
And,
tan θ = 2 tan (θ/2) / 1 - tan² (θ/2)
Substitute the value of θ here.
⟹ tan 330° = 2 tan (330/2)° / 1 - tan² (330/2)°
330° can be written as (360 - 30)°. so,
⟹ tan (360 - 30)° = 2 tan 165° / 1 - tan² 165°
- tan (360 - A) = - tan A
⟹ - tan 30° = 2 tan 165° / 1 - tan² 165°
- tan 30° = 1/√3
⟹ ( - 1/√3) (1 - tan² 165°) = 2 tan 165°
⟹ (- 1 + tan² 165°) / √3 = 2 tan 165°
⟹ tan² 165° - 1 = 2√3 tan 165°
⟹ tan² 165° - 2√3 tan 165 - 1 = 0
By splitting the middle term,
⟹ tan² 165° - (√3 + 2) tan 165° - (√3 - 2) tan 165° - 1 = 0
[ (√3 + 2) + (√3 - 2) = 2√3 & (√3 - 2)(√3 +
2) = (√3)² - (2)² = 3 - 4 = - 1 ]
⟹ tan 165° [ tan 165° - (√3 + 2) ] - (√3 - 2) [ tan 165° - (√3 + 2) ] = 0
⟹ [ tan 165° - (√3 + 2) ][ tan 165° - (√3 - √2) ] = 0
⟹ tan 165° = ( √3 + 2 , √3 - 2)
∴ tan (θ/2) = (√3 + 2 , √3 - 2).
Given :-
θ = 330
To Find :-
find the value of {tan(theta)/(2)}.
Solution :-
tanθ = 2 tan(θ/2)/1 - tan²(θ/2)
We know that
330 = (360 - 30)
(360 - θ) = -(tanθ)
tan(330) = 2 × (330/2)/1 - tan²(330/2)
→ 2 × tan 165/1 - tan²(165)
→ 2(tan 165)/1 - tan²(165)
As tan 30 = 1/√3
-(tan 30) = (-1/√3)
→ -(tan 30) = 2 tan165/1 - tan²165
→ -(tan 30)(1 - tan² 165) = 2 tan165
→ -(1/√3)(1 - tan²165) = 2 tan165
→ -1(1 - tan²165)/√3 = 2 tan165
→ -1 + tan²165 = √3(2 tan165)
→ -1 + tan²165 = 2√3 × tan165
Transposing all terms to LHS
→ -1 + tan²165 - 2√3 × tan165 = 0
→ tan²165 - 2√3 × tan165 - 1 = 0
On factorising
→ tan²165 - (2 + √3)tan165 - (√3 - 2)tan 165 - 1 = 0
→ tan 165[tan 165 - (2 + √3)] - tan165(√3 + 2) = 0
→ tan 165 = (2 + √3 & √3 - 2)
As
tan 165 = tan (330/2)
Hence
tan(θ/2) = (2 + √3 & √3 - 2)
Know More :-
In a right angled triangle
sin A = Opp/Hyp
cos A = Adj/Hyp
cot A = Adj/Opp
tan A = Opp/Adj
sec A = Hyp/Adj
cosec A = Hyp/Opp