Sin a/1+ cos a=cosec a - cot a – InterviewSolution

1.	Sin a/1+ cos a=cosec a - cot a
Answer» Question :- Prove that : $\implies\boxed{\sf{ \frac{ \sin a}{1 + \cos a} = \cosec a - \cot a }}\\$ Used Identity :- $\star \: \boxed{ \frac{ \cos \theta}{ \sin \theta} = \cot \theta \: } \\ \\ \star \: \boxed{ \frac{1}{ \sin \theta} = \cosec \theta} \\ \\ \star \: \boxed{ 1 - { \cos}^{2} \theta = { \sin}^{2} \theta}$ Explanation :- TAKE LHS ( left hand side ) $\to \sf{ \frac{ \sin a}{1 + \cos a} \: } \\ \\ \sf{ rationalise \: by \: 1 - \cos a} \\ \\ \to \sf{ \frac{ \sin a}{1 + \cos a} \times \frac{1 - \cos a}{1 - \cos a} } \\ \\ \to \sf{ \frac{ \sin a(1 - \cos a)}{(1 - \cos a)(1 + \cos a)} } \\ \: \\ \because \boxed{ \bf{ (x - y)(x + y) = {x}^{2} - {y}^{2} }} \\ \\ \to \sf{ \frac{ \sin a(1 - \cos a)}{1 - { \cos}^{2}a } } \\ \\ \to \sf{ \frac{ \sin a(1 - \cos a)}{ { \sin}^{2}a } } \\$ $\to \sf{ \frac{1 - \cos a}{ \sin a} } \: \\ \\ \to \sf{ \frac{1}{ \sin a} - \frac{ \cos a}{ \sin a} } \: \\ \\ \to \boxed{ \sf{ \cosec a - \cot a}}$ °•° LHS = RHS HENCE proved .

Sin a/1+ cos a=cosec a - cot a

Answer»

Question :-

Prove that :

$\implies\boxed{\sf{ \frac{ \sin a}{1 + \cos a} = \cosec a - \cot a }}\\$

Used Identity :-

$\star \: \boxed{ \frac{ \cos \theta}{ \sin \theta} = \cot \theta \: } \\ \\ \star \: \boxed{ \frac{1}{ \sin \theta} = \cosec \theta} \\ \\ \star \: \boxed{ 1 - { \cos}^{2} \theta = { \sin}^{2} \theta}$

Explanation :-

TAKE LHS ( left hand side )

$\to \sf{ \frac{ \sin a}{1 + \cos a} \: } \\ \\ \sf{ rationalise \: by \: 1 - \cos a} \\ \\ \to \sf{ \frac{ \sin a}{1 + \cos a} \times \frac{1 - \cos a}{1 - \cos a} } \\ \\ \to \sf{ \frac{ \sin a(1 - \cos a)}{(1 - \cos a)(1 + \cos a)} } \\ \: \\ \because \boxed{ \bf{ (x - y)(x + y) = {x}^{2} - {y}^{2} }} \\ \\ \to \sf{ \frac{ \sin a(1 - \cos a)}{1 - { \cos}^{2}a } } \\ \\ \to \sf{ \frac{ \sin a(1 - \cos a)}{ { \sin}^{2}a } } \\$

$\to \sf{ \frac{1 - \cos a}{ \sin a} } \: \\ \\ \to \sf{ \frac{1}{ \sin a} - \frac{ \cos a}{ \sin a} } \: \\ \\ \to \boxed{ \sf{ \cosec a - \cot a}}$