find the value of other trigonometric ratios, given that Sino m²-n²/ m²+n²
Answers
Step-by-step explanation:
Given :-
Sin θ = (m²-n²)/(m²+n²)
To find :-
Find the value of other trigonometric ratios ?
Solution :-
Given that
Sin θ = (m²-n²)/(m²+n²) ----------(1)
=> 1/Sin θ = 1/( (m²-n²)/(m²+n²)
=> Cosec θ = (m²+n²)/(m²-n²) --------(2)
On taking equation (1)
Sin θ = (m²-n²)/(m²+n²)
On squaring both sides then
=> (Sin θ)² = [(m²-n²)/(m²+n²)]²
=> Sin² θ = (m²-n²)²/(m²+n²)²
On subtracting the above equation from 1 then
=> 1- Sin² θ = 1-[ (m²-n²)²/(m²+n²)²]
=> Cos² θ = [(m²+n²)²-(m²-n²)²]/(m²+n²)²
Since Sin² A + Cos²A = 1
=>Cos²θ=[(m⁴+2m²n²+n⁴)-(m⁴-2m²n²+n⁴)]/(m²+n²)²
=> Cos²θ = (m⁴+2m²n²+n⁴-m⁴+2m²n²-n⁴)/(m²+n²)²
=>Cos²θ = (2m²n²+2m²n²)/(m²+n²)²
=> Cos²θ = 4m²n²/(m²+n²)²
=> Cos θ =√[4m²n²/(m²+n²)²]
=> Cos θ = √[(2mn)²/(m²+n²)²]
=> Cos θ = √[2mn/(m²+n²)]²
=> Cos θ = 2mn/(m²+n²) ---------(3)
=> 1/Cos θ = 1/[2mn/(m²+n²)]
=> Sec θ = (m²+n²)/(2mn) --------(4)
We know that
Tan θ = Sin θ / Cos θ
=> Tan θ = [ (m²-n²)/(m²+n²)] / [ 2mn/(m²+n²) ]
=> Tan θ = [ (m²-n²)/(m²+n²)] ×[(m²+n²) /(2mn)]
=> Tan θ = [ (m²-n²)(m²+n²)] / [(m²+n²)(2mn)]
=> Tan θ = (m²-n²)/(2mn) ---------(5)
=> 1/ Tan θ = 1/[(m²-n²)/(2mn)]
=> Cot θ = 2mn / (m²-n²) -----------(6)
Answer:-
The values of other trigonometric ratios are:
1) Cos θ = 2mn/(m²+n²)
2) Tan θ = (m²-n²)/(2mn)
3) Cosec θ = (m²+n²)/(m²-n²)
4) Sec θ = (m²+n²)/(2mn)
5) Cot θ = 2mn / (m²-n²)
Used formulae:-
→ Sin² A + Cos²A = 1
→ Tan θ = Sin θ / Cos θ
→ Sec θ = 1/ Cos θ
→ Cosec θ = 1/ Sin θ
→ (a+b)² = a²+2ab+b²
→ (a-b)² = a²-2ab+b²
→ (a/b)^n = a^n / b^n