If cos theta=x/y ,then tan theta +4/4 cot thata +1=
Answers
Given :-
cos θ = x/y
To find :-
The value of (tan θ + 4 )/(4 cot θ +1)
Solution :-
Given that
cos θ = x/y
=> 1 / cos θ = 1 / (x/y)
=> sec θ = y/x
On squaring both sides then
=> ( sec θ)² = (y/x)²
=> sec² θ = y²/x²
On subtracting 1 from both sides then
=> sec² θ - 1 = (y²/x²) - 1
=> sec² θ - 1 = (y²-x²)/x²
=> tan² θ = (y²-x²)/x²
Since, sec² A - tan² A = 1
=> tan θ = √[(y²-x²)/x²]
=> tan θ = √(y²-x²)/x
=> 1 / tan θ = 1/[√(y²-x²)/x]
=> cot θ = x/√(y²-x²)
Now,
The value of tan θ + 4
= [ √(y²-x²)/x] +4
= [√(y²-x²)+4x]/x
and
The value of 4 cot θ + 1
=> (4/ tan θ)+1
=>
= 4 [ x / √(y²-x²)]+1
=> [4x/√(y²-x²)]+1
=> [4x+√(y²-x²)]/√(y²-x²)
Now,
The value of (tan θ + 4 )/(4 cot θ +1)
= [{√(y²-x²)+4x}/x] / [{4x+√(y²-x²)}/√(y²-x²)]
= [{√(y²-x²)+4x}{√(y²-x²)}] / [x{4x+√(y²-x²)}]
= √(y²-x²)/x
Alternative Method :-
Given that
cos θ = x/y
=> 1 / cos θ = 1 / (x/y)
=> sec θ = y/x
On squaring both sides then
=> ( sec θ)² = (y/x)²
=> sec² θ = y²/x²
On subtracting 1 from both sides then
=> sec² θ - 1 = (y²/x²) - 1
=> sec² θ - 1 = (y²-x²)/x²
=> tan² θ = (y²-x²)/x²
Since, sec² A - tan² A = 1
=> tan θ = √[(y²-x²)/x²]
=> tan θ = √(y²-x²)/x
now ,
4 cot θ +1
=> (4/tan θ)+1
=>( 4+tan θ)/tan θ
Now,
(tan θ + 4 )/(4 cot θ +1)
=> (tan θ + 4 )/[(4+tan θ)/tan θ]
=> (tan θ + 4 ) tan θ/ (4+tan θ)
=> tan θ
Therefore,
(tan θ + 4 )/(4 cot θ +1) = tan θ
=> √(y²-x²)/x
Answer :-
The value of (tan θ + 4 )/(4 cot θ +1) is √(y²-x²)/x
Used formulae:-
★ sec θ = 1 / cos θ
★ cot θ = 1 / tan θ
★ sec² A - tan² A = 1