Question

1+cosa/sina-sina/1+cosa=2cota​

Answer 1

Question :-

Prove that :

$\longmapsto \: \frac{1 + \cos a}{ \sin a} - \frac{ \sin a}{1 + \cos a} = 2 \cot a$

Used Formula :-

$\star \: \: \boxed{ \sf{(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy} \\ \\ \star \: \: \boxed{ \sf \frac{ \cos \theta}{ \sin \theta} = \cot \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \star \: \: \boxed{ \: { \sin}^{2} \theta + { \cos}^{2} \theta = 1 \: \: \: \: \: \: \: \: \: \: \: \: }$

Explanation :-

Taking left hand side (LHS)

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

LHS (Left hand side)= RHS (Right hand side )

Hence proved .

Answer 2

$\dfrac{1+\cos A}{\sin A} -\dfrac{\sin A}{1+\cos A} =2\cot A$, proved

Step-by-step explanation:

To prove that, $\dfrac{1+\cos A}{\sin A} -\dfrac{\sin A}{1+\cos A} =2\cot A$.

L.H.S. = $\dfrac{1+\cos A}{\sin A} -\dfrac{\sin A}{1+\cos A}$

Taking LCM of denominator part, we get

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

Using the algebraic identity,

$(a+b)^{2} =a^{2} +2ab+b^2$

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

Using the trigonometric identity,

$\sin^2 A=1-\cos^2 A$

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

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

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

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

$=\dfrac{2\cos A}{\sin A}$

= $2\cot A$

= R.H.S., proved.

Thus, $\dfrac{1+\cos A}{\sin A} -\dfrac{\sin A}{1+\cos A} =2\cot A$, proved