Question

(SinA-cosA÷sinA+cosA)+(sinA+cosA÷sinA-cosA)=1÷(2sinA-1)

Answer 1

Solution

$\mathsf{L.H.S = \frac{sinA - cosA}{sinA + cosA} + \frac{sinA + cosA}{sinA - cosA}}$

By taking L.C.M we get that

$\mathsf{\Longrightarrow{\frac{{(sinA - cosA)}^{2} + {(sinA + cosA)}^{2} }{(sinA + cosA)(sinA - cosA)}}}$

On expanding 2sinA cosA cancels out and same terms add up.

$\mathsf{\Longrightarrow{\frac{2{sin}^{2}A + 2{cos}^{2}A }{sin^{2}A - cos^{2}A } = \frac{2({sin}^{2}A + {cos}^{2}A)}{{sin}^{2}A - 1 + sin^{2}A}}}$

Since sin²A + cos²A = 1 and cos²A = 1 - sin²A.

→ 2/(2 sin²A - 1) = R.H.S

Hence Proved