Question

SinA+cosA/sinA-cosA+sinA-cosA/sinA+cosA=2/sin^2A-1

Answer 1

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$\huge\bold\red{{solution:-}}$

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

By algebraic identities,

${(a+b)}^{2}+{(a-b)}^{2}=2{(a}^{2}+{b}^{2})$

$(a-b)(a+b)={a}^{2}-{b}^{2}$

${and}$

[By Trigonometric identity

${sin}^{2}A+{cos}^{2}A=1$

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

$=\frac{2}{sin^{2}A-(1-sin^{2}A)}$

By Trigonometric identity:

${cos}^{2}A=1-{sin}^{2}A$

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

$=\frac{2}{2(1-cos^{2}A)-1}\\=\frac{2}{2-2cos^{2}A-1}\\=\frac{2}{1-2cos^{2}A}---------(3)$

${Therefore},$

$\frac{sinA+cosA}{sinA-cosA}+\frac{sinA-cosA}{sinA+cosA}\\=\frac{2}{sin^{2}A-cos^{2}A}\\=\frac{2}{2sin^{2}A-1}\\=\frac{2}{1-2cos^{2}A}$

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