Question

1+cosA/sinA+sinA/1+cosA=2cosecA

Answer 1

Answer is in the image

Answer 2

Answer:

$\frac{1+cosA}{sinA}+\frac{sinA}{1+cosA}=2cosecA$

Step-by-step explanation:

$LHS=\frac{1+cosA}{sinA}+\frac{sinA}{1+cosA}\\=\frac{(1+cosA)^{2}+sin^{2}A}{sinA(1+cosA)}\\=\frac{1^{2}+cos^{2}A+2\times 1 \times cosA+sin^{2}A}{sinA(1+cosA)}$

\* By algebraic identity:

(a+b)² = a²+b²+2ab */

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

$=\frac{1+1+2cosA}{sinA(1+cosA)}$

/*By Trigonometric identity:

sin²A+cos²A = 1 */

$=\frac{2+2cosA}{sinA(1+cosA)}$

$=\frac{2(1+cosA)}{sinA(1+cosA)}$

$=\frac{2}{sinA}$

$=2cosecA$

$=RHS$

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