Question

secA-tanA/secA+tanA=cos^2A/(1+sinA)^2

Answer 1

It simple .go on making it

Answer 2

Answer:

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

Step-by-step explanation:

$LHS =\frac{secA-tanA}{secA+tanA}\\=\frac{(secA-tanA)(secA+tanA)}{(secA+tanA)(secA+tanA)}\\=\frac{sec^{2}A-tan^{2}A}{(secA+tanA)^{2}}$

/* By Trigonometric identity:

sec²A-tan²A = 1 */

$=\frac{1}{(secA+tanA)^{2}}$

$=\frac{1}{\left(\frac{1}{cosA}+\frac{sinA}{cosA}\right)^{2}}\\=\frac{1}{\left(\frac{(1+sinA)}{cosA}\right)^{2}}\\=\frac{1}{\frac{(1+sinA)^{2}}{cos^{2}A}}\\=\frac{cos^{2}A}{(1+sin^{2}A)}\\=RHS$

Therefore,

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

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