Question

prove that(1-sina+cosa)sq=2(1+cosa)(1-sina)​

Answer 1

Step-by-step explanation:

To prove --->

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

Proof --->

LHS = ( 1 - SinA + CosA )²

= { 1 + ( -SinA ) + CosA }²

We have an identity as follows

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

Applying it here we get

= (1)²+(-SinA)²+(CosA)²+2(1)Cos) +2(1)(-SinA) + 2 (-SinA ) CosA

= 1 + (Sin²A + Cos²A) - 2SinA + 2CosA

- 2SinA CosA

We have an identity

Sin²θ + Cos²θ = 1 , applying it here ,we get

= 1 + 1 - 2SinA + 2CosA - 2 SinA CosA

= 2 - 2 SinA + 2CosA - 2SinA CosA

Taking 2 , common from first two terms and 2CosA , from next two terms , we get

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

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

Taking 2 Common from second bracket , we get

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

= 2 ( 1 + CosA ) ( 1 - SinA ) = RHS

Additional information--->

(1) 1 + tan²θ = Sec²θ

(2) 1 + Cot²θ = Cosec²θ

Answer 2

Step-by-step explanation:

Consider the LHS:

 (1-sinA+cosA)2 = [(1-sinA) + cosA]2

 =  (1-sinA)2 + cos2A + 2(1-sinA)

cosA =  1 + sin2A − 2sinA + cos2A + 2(1-sinA)cosA  =  1 + (sin2A + cos2A) − 2sinA + 2(1-sinA)cosA  =  1 + 1 − 2sinA + 2(1-sinA)cosA    [Since, sin2A + cos2A =1] =  2 − 2sinA + 2(1-sinA)cosA =  2(1 − sinA) + 2(1-sinA)cosA =  2(1 − sinA)(1 + cosA) = RHS