prove that (1-cosA)(1+cosA)(1+cotA×cotA)=1
Answers
Answered by
13
Heya!!
---------
◾( 1 - CosA ) ( 1 + CosA) ( 1 + CotA × CotA) = 1
Now , using the identity ( a - b ) (a + b ) = ( a² - b² )
=>> ( 1 - Cos²A ) ( 1 + Cot²A)
=>> Now,
========
◀1 - Cos² A = Sin²A as
◀1 + Cot²A = Cosec²A
〰▪Putting in Values ,
( Sin²A ) ( Cosec²A)
=>>> Now, Sin²A = 1/ Cosec²A
We have !!
--------------
( Sin²A ) × 1/ Sin²A
= 1 = RHS!!
---------
◾( 1 - CosA ) ( 1 + CosA) ( 1 + CotA × CotA) = 1
Now , using the identity ( a - b ) (a + b ) = ( a² - b² )
=>> ( 1 - Cos²A ) ( 1 + Cot²A)
=>> Now,
========
◀1 - Cos² A = Sin²A as
◀1 + Cot²A = Cosec²A
〰▪Putting in Values ,
( Sin²A ) ( Cosec²A)
=>>> Now, Sin²A = 1/ Cosec²A
We have !!
--------------
( Sin²A ) × 1/ Sin²A
= 1 = RHS!!
Answered by
27
( 1 - CosA ) ( 1 + Cos A ) ( 1 + CotA × CotA ) = 1
LHS = ( 1 - Cos A ) ( 1 + Cos A ) ( 1 + CotA × Cot )
=> ( 1 - Cos²A ) ( 1 + Cot²A )
=> Sin²A × Cosec²A
=> Sin²A × 1/ Sin²A
=> 1
Hence,
LHS = RHS = 1.
LHS = ( 1 - Cos A ) ( 1 + Cos A ) ( 1 + CotA × Cot )
=> ( 1 - Cos²A ) ( 1 + Cot²A )
=> Sin²A × Cosec²A
=> Sin²A × 1/ Sin²A
=> 1
Hence,
LHS = RHS = 1.
Similar questions