Question

show that , sinA/1-cosA=cosecA+cotA​

Answer 1

${ \tt{ \underline{ \boxed{ \red{Trigonometry}}}}} \implies$

To Prove :

$\implies{ \sf{ \dfrac{SinA}{1 - Cos \: A}}}= { \sf{ Cosec \: A \: + Cot \: A}}$

Multiply and Divide the equation by ( 1 + Cos A )

$\implies{ \sf{ \dfrac{SinA}{1 - Cos \: A}}} \times { \sf\dfrac{ (\: 1 + \ \: Cos \: A) }{(1 + Cos \: A)}} \\ \\ \implies{ \sf{ \dfrac{SinA(1 + \: Cos \: A)}{1 {}^{2} - Cos {}^{2} \: A}}} \: \\ \\ \implies \: \: { \sf{ \dfrac{SinA +SinA \: Cos} {Sin {}^{2} A}}} \: \: \:$

[ Using Sin²∅ + Cos²∅ = 1 ]

★ Separate the terms ,

$\implies \: \: { \sf{ \dfrac{SinA } {Sin {}^{2} A}}} \: \: + { \sf{ \dfrac{SinA \: C os A }{Sin {}^{2} A}}} \:$

★ Sin A gets cancelled out both sides , we're left with

$\implies \: \: { \sf{ \dfrac{1 } {Sin A}}} \: \: + { \sf{ \dfrac{ \: Cos A }{Sin A}}} \:$

$\implies \: { \sf{Cosec A + Cot A}} \:$

Using ,

$\rightarrow \: { \boxed{ \bold{\dfrac{1}{Sin \theta} = Cosec \: \theta}}} \: \: and \: \: \rightarrow \: { \boxed{ \bold{\dfrac{Cos \theta}{Sin \theta} = Cot \: \: \theta}}} \: \: \:$

Hence ,

L.H.S = R.H.S

Answer 2

Answer:

hope that helps you

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