prove that root 1-cos a /1+cos a=1/cosec a + cot a
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√1+cosA/√1-cosA
Multiplying and dividing with conjugate of denominator
{√1+cosA/√1-cosA} × {√1+cosA/√1+cosA}
(√1+cosA)(√1+cosA) / (√1-cosA)(√1+cosA)
{√(1+cosA)(1+cosA)} / {√(1-cosA)(1+cosA)}
{√(1+cosA)²} / {√(1-cosA)(1+cosA)}
Since(a-b)(a+b) = a² - b²
Therefore (1-cosA)(1+cosA) = 1² - cos²A
(1+cosA) / √(1² - cos²A)
Now 1 - cos²A = sin²a
Thus (1+cosA) / √(sin²a)
(1+cosA) / sinA
Splitting the numerator
1/sinA + cosA/sinA
1/sinA = cosecA and cosA/sinA = cotA
Thus
1/sinA + cosA/sinA = cosecA + cotA
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