Question

(1)
1 + cos A - sin^2A/
sin A (1 + cosA)=
cot A​

Answer 1

To prove :-

$\bold{\frac{1 + \cos A - { \sin}^{2} A }{ \sin A(1 + \cos A)} = \cot A}$

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Consider LHS

$\bold{LHS = \: \frac{1 + \cos A - { \sin}^{2} A }{ \sin A(1 + \cos A)}}$

we know that,

• sin²A = 1 - Cos²A

$\bold{ \implies LHS = \: \frac{1 + \cos A - (1 - { \cos}^{2}A) }{ \sin A(1 + \cos A)}}$

and we know that,

• 1-cos²A = (1-cosA)(1+cosA)

$\bold{ \implies LHS = \: \frac{1 + \cos A - [(1 - \cos A)(1 + \cos A)]}{ \sin A(1 + \cos A)}}$

$\bold{ \implies LHS = \: \frac{1 + \cos A[1 - (1 - cosA)] }{ \sin A(1 + \cos A)}}$

$\bold{ \implies LHS = \: \frac{1 + \cos A( \cancel1 - \cancel 1 + cosA) }{ \sin A(1 + \cos A)}}$

$\bold{ \implies LHS = \: \frac{1 + \cos A( cosA) }{ \sin A(1 + \cos A)}}$

$\bold{ \implies LHS = \: \frac{ \cancel{(1 + \cos A)} cosA}{ \sin A \cancel{(1 + \cos A)}}}$

$\bold{ \implies LHS = \: \frac{cosA}{ \sin A}}$

and we know that,

$\boxed {\bold{\frac{ \cos A}{ \sin A} = \cot A}}$

$\bold{ \implies LHS = \cot A =RHS }$

hence, proved