if cot theta=√9-x^2, find sec theta tan theta
Answers
Given :-
cot θ = √(9-x²)
To find :-
• sec θ
• tan θ
Solution :-
Given that
cot θ = √(9-x²)
=> 1 / cot θ = 1/√(9-x²)
=> tan θ = 1/√(9-x²) --------(1)
On squaring both sides then
=> (tan θ)² = [1/√(9-x²)]²
=> tan² θ = 1/(9-x²)
On adding 1 both sides then
=> tan² θ + 1 = [1/(9-x²)] + 1
=> tan² θ + 1 = [1 + (9-x²)]/(9-x²)
=> tan² θ + 1 = (1+9-x²)/(9-x²)
=> tan² θ + 1 = (10-x²)/(9-x²)
We know that
sec² A - tan² A = 1
Therefore, sec² θ = (10-x²)/(9-x²)
=> sec θ = √[(10-x²)/(9-x²)]
Answer :-
• sec θ = √[(10-x²)/(9-x²)]
• tan θ = 1/√(9-x²)
Used formulae:-
• cot θ = 1 / tan θ
• sec² A - tan² A = 1
Step-by-step explanation:
Given :-
cot θ = √(9-x²)
To find :-
• sec θ
• tan θ
Solution :-
Given that
cot θ = √(9-x²)
=> 1 / cot θ = 1/√(9-x²)
=> tan θ = 1/√(9-x²) --------(1)
On squaring both sides then
=> (tan θ)² = [1/√(9-x²)]²
=> tan² θ = 1/(9-x²)
On adding 1 both sides then
=> tan² θ + 1 = [1/(9-x²)] + 1
=> tan² θ + 1 = [1 + (9-x²)]/(9-x²)
=> tan² θ + 1 = (1+9-x²)/(9-x²)
=> tan² θ + 1 = (10-x²)/(9-x²)
We know that
sec² A - tan² A = 1
Therefore, sec² θ = (10-x²)/(9-x²)
=> sec θ = √[(10-x²)/(9-x²)]
Answer :-
• sec θ = √[(10-x²)/(9-x²)]
• tan θ = 1/√(9-x²)
Used formulae:-
• cot θ = 1 / tan θ
• sec² A - tan² A = 1