Question

Prove the following identity :
(1 + cot A - cosec A) (1 + tan A + sec A)=2
​

Answer 1

LHS =

$(1 + cot \: a - cosec \: a) (1 + tan \: a + sec \: a)=2$

$(1 + \frac{ \cos(a) }{ \sin(a) } - \frac{1}{ \sin(a) } )(1 + \frac{ \sin(a) }{cos(a)} + \frac{1}{ \cos(a) } )$

Taking LCM

$( \frac{ \sin(a) + \cos(a) - 1 }{ \sin(a) } )( \frac{ \cos(a) + \sin(a) + 1) }{ \cos(a) } )$

$\frac{ {( \sin(a) + \cos(a)) }^{2} - {1}^{2} }{ \sin(a) \cos(a) }$

$\frac{ { \sin }^{2}a + { \cos }^{2}a \: + 2 \sin(a) \cos(a) - 1 }{ \sin(a) \cos(a) }$

$\frac{1 - 2 \sin(a) \cos(a) - 1 }{ \sin(a) \cos(a) }$

$\frac{2 \sin(a) \cos(a) }{ \sin(a) \cos(a) }$

2 = RHS

LHS=RHS

Answer 2

SOLUTION:-

We have,

(1+cotA -cosecA) (1+tanA+secA)=2

Take L.H.S

$(1 + tanA+ secA)(1 + cotA - cosecA) \\ \\ = > (1 + cotA - cosecA) + tanA(1 + cotA - cosecA) + secA( 1 + cotA - cosecA) \\ \\ = > (1 + cotA - cosecA) + tanA + tanA\: cotA- tanA \: cosecA + secA + secA\: cotA - secA\: cosecA \\ \\ = > 1 + cotA - cosecA + tanA + 1 - tanA \: cosecA+ secA + \frac{1}{cosA} \times \frac{cosA}{sinA} - secA \: cosecA \\ \\ = > 2 + cotA - cosecA+ tanA - tanA\: cosecA + secA + cosecA - secA \: cosecA\\ \\ = > 2 + (cotA+ tanA) - (tanA \: cosecA - secA) - secA\: cosecA\\ \\ = > 2 + ( \frac{cosA}{sinA } + \frac{sinA}{cosA} ) - ( \frac{sinA}{cosA} \times \frac{1}{sinA} - \frac{1}{cosA} ) - secA \: cosecA\\ \\ = > 2 + \frac{ {cos}^{2}A + {sin}^{2} A}{sinA \: cosA} - ( \frac{1}{cosA} - \frac{1}{cosA} ) - secA \: cosecA \\ \\ = > 2 + secA \: cosecA - secA\: cosecA \\ \\ = > 2 + 0 \\ \\ = > 2 \: \: \: \: \: \: \: \: \: \: \: [R.H.S.]$

Proved.