sinA+cosA=p ,secA+cosecA=q (cosA)^2/[2-2(sinA)^2-(cosecA)^2]=p^2/(2(p^2)-q^2
To solve this equation, let's simplify each side step by step.
Starting with the left side,
(cosA)^2/[2-2(sinA)^2-(cosecA)^2]
We can simplify the denominator using trigonometric identities:
2 - 2(sinA)^2 - (cosecA)^2
Recall that cosecA is the reciprocal of sinA, so we can rewrite it as:
2 - 2(sinA)^2 - (1/(sinA)^2)
Now, let's simplify it further:
2 - (2(sinA)^2 + 1/(sinA)^2)
Multiplying through by (sinA)^2, we get:
2(sinA)^2(sinA)^2 - 2(sinA)^2 - 1
Let's denote (sinA)^2 as x for simplicity:
2x^2 - 2x - 1
Now, moving on to the right side of the equation:
p^2/(2(p^2) - q^2)
We can substitute p = sinA + cosA and q = secA + cosecA:
(p = sinA + cosA) -> p^2 = (sinA + cosA)^2 = (sinA)^2 + 2sinAcosA + (cosA)^2
(q = secA + cosecA) -> q^2 = (secA + cosecA)^2 = (secA)^2 + 2secAcosecA + (cosecA)^2
Now, let's substitute these values:
p^2/(2(p^2) - q^2) = [(sinA)^2 + 2sinAcosA + (cosA)^2]/[2((sinA)^2 + 2sinAcosA + (cosA)^2) - ((secA)^2 + 2secAcosecA + (cosecA)^2)]
Simplifying further:
[(sinA)^2 + 2sinAcosA + (cosA)^2]/[2(sinA)^2 + 4sinAcosA + 2(cosA)^2 - (secA)^2 - 2secAcosecA - (cosecA)^2]
Using trigonometric identities:
[(sinA)^2 + 2sinAcosA + (cosA)^2]/[(sinA)^2 + (cosA)^2 - 1 - 1/(sinA)^2 - 1/(cosA)^2]
Now, let's factor out the numerator:
[(sinA + cosA)^2]/[(sinA)^2 + (cosA)^2 - 1 - 1/(sinA)^2 - 1/(cosA)^2]
Substituting the values of p and q:
[(p)^2]/[(sinA)^2 + (cosA)^2 - 1 - 1/(sinA)^2 - 1/(cosA)^2]
Simplifying further:
[(p)^2]/[1 - (1 - (sinA)^2 - (cosA)^2) - 1/(sinA)^2 - 1/(cosA)^2]
Recall that sin^2(A) + cos^2(A) = 1:
[(p)^2]/[1 - 0 - 1/(sinA)^2 - 1/(cosA)^2]
Now, we have the right side of the equation simplified. We can put everything together:
2x^2 - 2x - 1 = [(p)^2]/[1 - 0 - 1/(sinA)^2 - 1/(cosA)^2]
This is the simplified form of the equation. To find the solution, you can further manipulate this equation, possibly using the quadratic formula, to solve for x.