Question

( 1+cot x–cosec x ) (1+tan x +sec x) =2

Answer 1

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Answer 2

Answer:

To show: ( 1 + cot x - cosec x )( 1 + tan x + sec x ) = 2

Consider,

LHS = ( 1 + cot x - cosec x )( 1 + tan x + sec x )

$=(1+\frac{sin\,x}{cos\,x}+\frac{1}{cos\,x})(1+\frac{cos\,x}{sin\,x}-\frac{1}{sin\,})$

$=(\frac{cos\,x+sin\,x+1}{cos\,x})(\frac{sin\,x+cos\,x-1}{sin\,x})$

$=(\frac{(cos\,x+sin\,x)+1}{cos\,x})(\frac{(cos\,x+sin\,x)-1}{sin\,x})$

Using, ( a - b )( a + b ) = a² - b²

$=\frac{(cos\,x+sin\,x)^2-1^2}{cos\,x\:sin\,x}$    

Using, ( a + b )² = a² + b² + 2ab

$=\frac{cos^2\,x+sin^2\,x+2\:cos\,x\:sin\,x-1}{cos\,x\:sin\,x}$

$=\frac{1+2\:cos\,x\:sin\,x-1}{cos\,x\:sin\,x}$

$=\frac{2\:cos\,x\:sin\,x}{cos\,x\:sin\,x}$

$=2$

$=RHS$

Hence Proved.