sec o=17/8 find 3-4 sin2 o/4cos2tita-3
Answers
Step-by-step explanation:
Given,
secA = 17/8
To Find :-
Value of :-
3 - 4 sin²A/4cos²A - 3
How To Do :-
Here they given the value of 'secA' and we are asked to find the value of 3-4sin²A/4cos²A - 2. So by applying Pythagoras theorem we need to find the value of another side . After obtaining that we need to find the value of 'sinA' and 'cosA'.
Formula Required :-
Pythagoras theorem :-
(hypotenuse side)² = (opposite side)² + (adjacent side)²
secA = hypotenuse side/adjacent side
sinA = opposite side/hypotenuse side
cosA = 1/secA
Solution :-
secA = 17/8
hypotenuse side/adjacent side = 17/8
→ hypotenuse side = 17
Adjacent side = 8
Let,
opposite side be 'x'
Applying Pythagoras theorem :-
(17)² = (8)² + (x)²
289 = 64 + x²
289 - 64 = x²
225 = x²
x = √225
x = 15
Opposite side = 15
→ cosA = 1/secA
= 1/17/8
= 8/17
sinA = opposite side/hypotenuse side
= 15/18.
3 - 4sin²A/4cos²A - 3
= 3 - 4(15/17)²/4(7/17)² - 3
= 3 - 4(225/289)/4(49/289)-3
= 3 - 900/289/196/289 + 3
= 867-900/289/196+867/289
= -33/289/1063/289
= -33/1063