SinA+cosA/sinA-cosA+sinA-cosA/sinA+cosA=2/sin^2A-cos^2A

Question

Trigonometry of std 10

Answer 1

I believe your problem is missing parentheses.

Without parentheses, following the order of operations rules, this is what happens to your equation:

sin A + (cos A/ sin A) - cos A + sin A - (cos A / sin A) + cos A = 2 sin A ≠ (2 / sin² A) - cos²A

With parentheses:

[(sin A + cos A) / (sin A - cos A)] + [(sin A - cos A) / (sin A + cos A)] = 2 / (sin² A - cos² A)

To combine fractions, you must have a common denominator. The common denominator of these two fractions is [(sin A + cos A) * (sin A - cos A)] = sin² A - cos² A

To get the common denominator for the first fraction, multiply numerator and denominator by (sin A + cos A). The first fraction becomes:

[(sin A + cos A)² / (sin² A - cos² A)]

To get the common denominator for the second fraction, multiply numerator and denominator by (sin A - cos A). The second fraction becomes:

[(sin A - cos A)² / (sin² A - cos² A)]

Now that the fractions have the common denominator, we can work with the numerators and combine them.

Multiply out the numerators:

(sin A + cos A)² = sin² A + 2 (sin A)(cos A) + cos² A

(sin A - cos A)² = sin² A - 2 (sin A)(cos A) + cos² A

Add numerators: the cross-terms cancel, answer reduces to:

2 sin² A + 2 cos² A = 2(sin² A + cos² A)

sin² A + cos² A = 1 (trig identity) --> 2(sin² A + cos² A) = 2*1 = 2

Combine numerator and denominator:

[(sin A + cos A) / (sin A - cos A)] + [(sin A - cos A) / (sin A + cos A)] = 2 / (sin² A - cos² A) Q.E.D.