Question

Please answer fast ​

Answer 1

Answer:

hope it helps you.....

Answer 2

To prove --->

$\dfrac{1 \: + cos \: \alpha }{sin \: \alpha } \: - \dfrac{sin \alpha }{1 \: + \: cos \: \alpha } \: = 2 \: cot \: \alpha$

Concept used ---->

1)

$\green{{(a \: + \: b \: )}^{2} \: = {a}^{2} \: + {b}^{2} \: + \: 2 \: a \: b} }$

2)

${sin}^{2} \alpha \: = \: 1 \: - \: {cos}^{2} \: \alpha$

3)

$cot \alpha \: = \dfrac{cos \: \alpha }{sin \: \alpha }$

Proof----> LHS

$= \dfrac{1 \: + \: cos \: \alpha }{sin \: \alpha } \: - \: \dfrac{sin \alpha }{1 \: + \: cos \: \alpha }$

$= \dfrac{( \: 1 \: + \: cos \alpha \: ) ^{2} \: - {sin}^{2} \: \alpha }{sin \: \alpha \: ( \: 1 \: + \: cos \alpha \: ) }$

$= \dfrac{1 \: + \: {cos}^{2} \alpha \: + \: 2 \: cos \alpha \: - \: {sin}^{2} \alpha }{sin \alpha \: ( \: 1 \: + \: cos \alpha \: )}$

$= \dfrac{1 \: + \: {cos}^{2} \alpha \: + \: 2 \: cos \alpha \: - \: ( \: 1 \: - \: {cos}^{2} \alpha \: ) }{sin \alpha \: ( \: 1 \: + \: cos \alpha \: )}$

$= \dfrac{1 \: + {cos}^{2} \alpha \: + 2 \: cos \alpha \: - \: 1 \: + \: {cos}^{2} \alpha }{sin \alpha \: ( \: 1 \: + \: cos \alpha \: )}$

$- 1 \: and \: + 1 \: cancel \: out \: from \: numerator \: and \: we \: get$

$= \dfrac{2 \: cos \alpha \: + \: 2 \: {cos}^{2} \alpha }{sin \alpha \: ( \: 1 \: + \: cos \alpha \: )}$

$= \dfrac{2 \: cos \alpha \: ( \: 1 \: + \: cos \alpha \: )}{sin \alpha \: ( \: 1 \: + \: cos \alpha \: )}$

$( \: 1 \: + \: cos \alpha \: ) \: is \: cancel \: out \: from \: numerator \: and \: denominator \: and \: we \: get \:$

$= \dfrac{2 \: cos \alpha }{sin \alpha }$

$= 2 \: ( \dfrac{cos \alpha }{sin \alpha \: } )$

$= 2 \: cot \alpha$

= RHS