Math, asked by hshamimara2077, 1 year ago

(SinA/1-cosA - 1-cosA/SinA)(cosA/1-sinA - 1-sinA/cosA) = 4

Answers

Answered by ilikeme
27

see the draft below for the answer

i hope it helps u

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Ayshasyshu: Thankzz
Answered by amitnrw
6

proved that ( SinA/(1 - CosA)  -  (1 - CosA)/SinA ) ( CosA/(1 - SinA)  - (1 - SinA)/CosA)  = 4

Step-by-step explanation:

to be proved

( SinA/(1 - CosA)  -  (1 - CosA)/SinA ) ( CosA/(1 - SinA)  - (1 - SinA)/CosA)  = 4

LHS  = ( SinA/(1 - CosA)  -  (1 - CosA)/SinA ) ( CosA/(1 - SinA)  - (1 - SinA)/CosA)

SinA/(1 - CosA)  -  (1 - CosA)/SinA

= (Sin²A - (1 - CosA)²)/(1 - CosA)(SinA)

= (Sin²A - (1 + Cos²A - 2CosA) / (1 - CosA)(SinA)

=  (Sin²A - 1 - Cos²A + 2CosA) / (1 - CosA)(SinA)

putting 1 = Sin²A + Cos²A

= (Sin²A - Sin²A - Cos²A - Cos²A + 2CosA) / (1 - CosA)(SinA)

= (2CosA - 2Cos²A)/ (1 - CosA)(SinA)

= 2CosA(1 - CosA) / (1 - CosA)(SinA)

= 2CosA/SinA

CosA/(1 - SinA)  - (1 - SinA)/CosA

= (Cos²A - (1 - SinA)²)/(1 - SinA)CosA

= (Cos²A - 1 - Sin²A + 2SinA)/(1 - SinA)CosA

= ( Cos²A - Cos²A - Sin²A - Sin²A + 2SinA)/(1 - SinA)CosA

= 2SinA(1 - SinA)/(1 - SinA)CosA

= 2SinA/CosA

LHS = (2CosA/SinA)( 2SinA/CosA)

= 4

= RHS

QED

proved

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