Given: secA tanA = x

To Find: sinA

Explanation:
We know that:
- secA = 1/cosA
- tanA = sinA/cosA

Substituting the value of secA and tanA in the given equation, we get:
1/cosA * sinA/cosA = x
sinA/cos²A = x
sinA = x cos²A

Now, we need to find the value of cos²A. For that, we can use the identity:
cos²A + sin²A = 1
cos²A = 1 - sin²A

Substituting this value in the earlier equation, we get:
sinA = x(1 - sin²A)
sin³A + x sinA - x = 0

This is a cubic equation in sinA. We can solve it using the cubic formula, but that would be quite tedious. Instead, we can use the rational root theorem to find one of the roots and then factorize the cubic equation.

The rational root theorem states that if a polynomial with integer coefficients has a rational root, then that root must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

In our case, the constant term is -x and the leading coefficient is 1. So, the possible rational roots are:
±1, ±x, ±1/2, ±x/2, ±1/3, ±x/3, ...

We can try these values one by one until we find a root. Let's say we find a root as sinA = p/q. Then, we can divide the cubic equation by sinA - p/q to get a quadratic equation, which can be easily solved using the quadratic formula.

Once we find the roots of the equation, we can use them to find the value of sinA.

Example:
Let x = 2. Then, we need to solve the equation:
sin³A + 2 sinA - 2 = 0

Possible rational roots are:
±1, ±2, ±1/2, ±1, ±1/3, ±2/3, ±2/5, ±4/5, ...

We can try sinA = 1, sinA = -1, sinA = 2/3, sinA = -2/3, sinA = 2/5, and sinA = -4/5. Out of these, sinA = 2/5 is a root.

Dividing the cubic equation by sinA - 2/5, we get:
sin²A + (2/5)sinA + 1 = 0

Using the quadratic formula, we get:
sinA = (-2/5 ± √(4/25 - 4))/2
sinA = (-2/5 ± √(4/25 - 4))/2

We can ignore the negative root as sinA cannot be negative. So, the value of sinA is:
sinA = (-2/5 + √(4/25 - 4))/2
sinA ≈ 0.584

Therefore, sinA ≈ 0.584 when secA tanA = 2

Sec^2A-tan^2A =1(secA+tanA)(secA-tanA)=1x.(secA-tanA)=1(secA-tanA)=1/x eq isecA + tanA = x eqiiadding both the equation we get2 secA =1+x^2/xsecA=1+x^2/2xtanA=x^2-1/2xsinA =tanA/secA=x^2-1/x^2+1TATH'S ALL