Show that: sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) = 1.

Solution:

To prove sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) = 1, we will use trigonometric identities of trigonometry.

LHS = sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) 

= sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) = 1sin A / [ (1/ cos A) + (sin A / cos A) - 1] + cos A / [ (1/ sin A) + (cos A / sin A) - 1]

On solving this by taking LCM, we get 

= sin A / [( 1 + sin A - cos A ) / cos A ] + [ cos A / ( 1 + cos A - sin A ) / sin A ]

= sin A × cos A [ ( 1 / 1 + sin A - cos A ) + ( 1 / 1 + cos A - sin A ) ]

By taking LCM and solving this, we get

⇒ sin A × cos A [ ( 2 ) / 1 + (sin A - cos A ) ( 1 - ( sin A + cos A ) ]

Using ( a - b ) ( a + b ) = a² - b² we will solve this.

⇒ sin A × cos A / [ 1 + (sin A - cos A ) ( 1 - ( sin A + cos A ) 

⇒ 2 sin A × cos A / (1 - sin² A + cos² A - 2 sin A × cos A)

By using sin² A + cos² A = 1, we get

⇒ 2 sin A × cos A / 2 sin A × cos A 

= 1

= RHS

Hence Proved that sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) = 1

Thus, sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) is equal to 1.

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Show that: sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) = 1.

Summary:

sin A / (sec A + tan A - 1) + cos A / (cosec A + cot A - 1) = 1 can be proved using identities.