if secA-tanA =10 than sinA =?,cosA =?,secA =?
use sec²A=1+tan²A
and let t=tan(A)
sec(A)-tan(A)=10
√(1+t^2)-t=10
√(1+t^2)=(t+10)²
1=20t+100
t=-99/20
A=atan(-99/20)=-78.5788 (approx.)
sec(A)=1/√(1+t^2)=101/20
cos(A)=1/sec(A)=20/101
sin(A)=-sqrt(1-(20/101)^2)=99/101
(note: A is in Q4, so sin(A)<0)
To find the values of sinA, cosA, and secA when secA - tanA = 10, we can use some trigonometric identities and equations.
First, let's rearrange the given equation: secA - tanA = 10
Since secA = 1/cosA and tanA = sinA/cosA, we can substitute these values into the equation:
1/cosA - sinA/cosA = 10
Combining the fractions on the left side of the equation, we have:
(1 - sinA) / cosA = 10
Now, let's cross-multiply:
1 - sinA = 10 * cosA
Next, let's square both sides of the equation to eliminate the square root:
(1 - sinA)^2 = (10 * cosA)^2
Expanding both sides of the equation:
1 - 2sinA + sin^2A = 100cos^2A
Using the Pythagorean identities sin^2A + cos^2A = 1, we replace sin^2A and cos^2A with 1 - cos^2A:
1 - 2sinA + (1 - cos^2A) = 100cos^2A
Rearranging and combining like terms:
2cos^2A + 2sinA - 2 = 0
Dividing by 2:
cos^2A + sinA - 1 = 0
Now, we can use the quadratic formula to solve for cosA:
cosA = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 1, and c = -1. Substituting these values into the formula:
cosA = (-(1) ± √((1)^2 - 4(1)(-1))) / 2(1)
cosA = (-1 ± √(1 + 4)) / 2
cosA = (-1 ± √5) / 2
Therefore, cosA = (-1 + √5)/2 or cosA = (-1 - √5)/2
Once we have the values of cosA, we can find the values of sinA and secA using the trigonometric identities:
sinA = √(1 - cos^2A)
secA = 1/cosA
Substituting the values of cosA:
sinA = √(1 - ((-1 + √5)/2)^2)
secA = 1 / ((-1 + √5)/2)
By simplifying these expressions, we can find the exact values of sinA and secA.