Question

sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA= 2/sin²A-cos²A=2/1-2cos ²A,Prove it by using trigonometric identities.

Answer 1

Given :

Sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA= 2/sin²A-cos²A=2/1-2cos ²A

LHS = Sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA

= {(Sin A+cosA)² + (Sin A - cosA)²} /(sin A -cosA)(sinA+cosA)

={ (sin²A+cos²A + 2sinA.cosA) + (sin²A+cos²A - 2sinA.cosA)} / (sin² A - cos²A)

= {(1+2sinA.cosA) + (1- 2sinA.cosA) / (sin² A - cos²A)

[sin²θ+cos²θ=1]

= 1 + 1 / (sin² A - cos²A)

LHS = 2 / (sin² A - cos²A) = 2/(1- cos²A - cos²A) = 2/ (1- 2cos²A) = RHS

[sin²θ = 1 - cos²θ]

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Hi ,

*******************************

We know that ,

1 ) ( a + b )² + ( a - b )² = 2( a² + b² )

2 ) ( a + b )( a - b ) = a² - b²

3 ) sin² A + cos² A = 1

4 ) sin² A - 1 = cos² A
*****************************************

i ) ( sinA+cosA)/(sinA-cosA)+(sinA-cosA)/(sinA+cosA )

= [(sinA+cosA)²+(sinA-cosA)²]/[(sinA-cosA)(sinA+cosA)]

=[2(sin²A + cos²A)]/(sin²A - cos²A )

= 2/( sin²A - cos²A ) ---( 1 )

= 2/[ 1 - cos²A - cos²A ]

= 2/( 1 - cos²A ) ---( 2 )

I hope this helps you.

: )