Question

prove this identity cos A/1+sin A + 1+sinA/cosA = 2secA​

Answer 1

Answer:

PROVED

Step-by-step explanation:

Taking LCM and adding the 2 fractions:

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

$[$ opening $(1 + sinA)^2$ as $(a+b)^2 = a^2 + 2ab + b^2]$

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

Since $cos^2A + sin^2A = 1$

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

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

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

So, now (1 + sinA) in the numerator and denominator gets cancelled.

= $\frac{2}{cosA}$

= $2secA$                      [since $\frac{1}{cos} = secA$ ]

Hence Proved

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