Question

anyone solve this question
and give solutions​

Answer 1

( 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.