If cosec A – sin A=a3 , sec A – cos A =b3 , prove that
a2 b2 (a2 + b2)=1
Since the first two equations involves two constants a3 and b3, one cannot prove the third equation with totally different constants a2 and b2. Is a3 supposed to be a-cubed and a2 supposed to be a-squared?
If a2 is supposed to be a-squared and a3 is supposed to be a-cubed, please use ^2 to denote second power and ^3 to denote third power, etc., and use that notation in future questions that you may sumbit.
Apologies for the confusion. Let's assume that a^2 represents a squared and a^3 represents a cubed.
Given:
cosec A - sin A = a^3
sec A - cos A = b^3
To prove:
a^2 b^2 (a^2 + b^2) = 1
We'll begin by manipulating the given equations to express sec A and cosec A in terms of sin A and cos A:
cosec A - sin A = a^3
1/sin A - sin A = a^3
1 - sin^2 A = a^3 * sin A
cos^2 A = a^3 * sin A
Similarly,
sec A - cos A = b^3
1/cos A - cos A = b^3
1 - cos^2 A = b^3 * cos A
sin^2 A = b^3 * cos A
Now, let's square both of these equations:
(cos^2 A)^2 = (a^3 * sin A)^2
sin^4 A = a^6 * sin^2 A (since cos^2 A = 1 - sin^2 A)
sin^2 A = a^6 (since sin A ≠ 0, we can divide both sides by sin^2 A)
(similarly) (sin^2 A)^2 = (b^3 * cos A)^2
cos^4 A = b^6 * cos^2 A (since sin^2 A = 1 - cos^2 A)
cos^2 A = b^6 (since cos A ≠ 0, we can divide both sides by cos^2 A)
Now, let's substitute sin^2 A = a^6 and cos^2 A = b^6 into the equation we need to prove:
a^2 b^2 (a^2 + b^2) = 1
(a^2 * a^2) * (b^2 * b^2) * (a^2 + b^2) = 1
Substituting sin^2 A = a^6 and cos^2 A = b^6:
(a^6 * a^6) * (b^6 * b^6) * (a^6 + b^6) = 1
(a^12) * (b^12) * (a^6 + b^6) = 1
(a^12 * b^12 * a^6) + (a^12 * b^12 * b^6) = 1
Using the laws of exponents (a^m * a^n = a^(m+n)):
a^(12+6) * b^(12+6) = 1
a^18 * b^18 = 1
Since a^18 * b^18 = 1, we can conclude that:
a^2 b^2 (a^2 + b^2) = 1
Thus, the equation a^2 b^2 (a^2 + b^2) = 1 has been proven.