Question

2) सिद्ध करा- (sec∅ - cos∅) (cot∅ + tan∅) = tan∅.sec∅

ans:​

Answer 1

(sec∅ - cos∅) (cot∅ + tan∅) = tan∅.sec∅

LHS

(1/cos∅ - cos∅)(1/tan∅ + tan∅)

(1 - cos²∅)/(cos∅) * (1 + tan²∅)/tan∅

Itnai mai toh kuch nhi batana tha -°-

(sin²∅/cos∅)(sec²∅/tan∅)

sin∅/cos²∅

tan∅/cos∅ => tan∅sec∅

Manje prove.... xD xD

Shayd yhi bolte hai na

Answer 2

$(sec∅ - cos∅) (cot∅ + tan∅) = tan∅ \: . \: sec∅$

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

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

$= \frac{ { \sin }^{2} ∅ }{ \cos∅ } \times \frac{1}{ \sin ∅ \cos ∅ }$

$= \frac{ \sin∅ }{ \cos∅ } \times \frac{1}{ \cos∅ }$

$= \tan∅ cos ∅$

$L.H.S = R.H.S$